Accenture Latest Coding Questions with Solutions

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Accenture Coding Question 2022: 

Accenture is a global professional services company with leading capabilities in digital, cloud and security. Here we discuss some Accenture Coding Questions.
 
Innovative work. A unique, diverse workplace environment. Continuous learning and development opportunities. These are just some of the reasons we're consistently recognized as one of the best companies to work for, and why our people choose to grow careers at Accenture.
 
Accenture offers exceptional development opportunities and an attractive rewards package, in addition to a competitive salary. It's the ideal start to your career.

The Accenture placement test has 4 sections with a total of 112 questions.The sections are:

Cognitive Ability: 50 questions in 50 minutes. There are three sections under this:

  • Verbal Ability
  • Analytical Reasoning
  • Numerical Ability

Technical Assessment: 40 questions in 40 minutes. There are questions from:

  • Fundamental of Networking Security and Cloud.
  • Pseudo Code
  • Common application and MS Office.

Coding test: 2 questions in 45 minutes.

Communication: 20 questions in 20 minute

Here in this blog we can discuss some :

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Here is some tricks to crack Accenture and companies like Accenture also interview tricks Interview Preparation. study about that company and their profile and statistics. focus on basic questions that are useful to them. Download Study Material for AMCAT Exam Preparation.

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Some FAQ

How hard is the Accenture coding test?
Accenture has recently introduced a few changes in its coding test. The difficulty of this section lies between medium to high.
Accenture has three rounds. Before these four rounds, Accenture campus placements are not that difficult to crack and with consistent effort, they can be cracked easily. If one can learn about what the four rounds are, cracking campus placements can be smooth.

Accenture Coding Question With Solutions :

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Accenture Question 1 : Longest Decreasing Sub-sequence (LDS)

Given an integer array 'A' (1<= length(A) <= 1000), find the length of its Longest Decreasing Sub-sequence (LDS). The elements of an LDS are sorted in monotonic/Strict decreasing order.
A Sequence is a particular order in which related objects follow each other. A sub- sequence is a sequence obtained by omitting some of the elements of a large sequence.
You need to fill in a function that takes two inputs0- integer 'n' and an integer array 'A' containing 'n' integers and return the length of LDS.

Input Specification:
input1 : integer n denoting size of array
input2 : integer array 'A' containing n elements

Example 1:
input1 : 3
input2 : {1,3,2}
Output : 2
Explanation
{3,2} is the longest decreasing sub-sequence of {1,3,2} . Hence the elements count in LDS is 2

Example 2:
input1: 5
input2: {41,18467,6334,26500,19169}
Output: 2
Explanation
{26500, 19169} is the longest decreasing sub-sequence of {41,18467,6334,26500,19169}, Hance the elements counts in LDS is 2.
import java.util.Scanner;
public class LongestDecresing {
  static int lds(int arr[], int n) {
    int lds[] = new int[n];
    int i, j, max = 0;
    for (i = 0; i < n; i++)
      lds[i] = 1;
    for (i = 1; i < n; i++)
      for (j = 0; j < i; j++)
        if (arr[i] < arr[j] && lds[i] < lds[j] + 1)
          lds[i] = lds[j] + 1;
    for (i = 0; i < n; i++)
      if (max < lds[i])
        max = lds[i];
    return max;
  }

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int[] arr = new int[n];
    for (int i = 0; i < n; i++) {
      arr[i] = sc.nextInt();
    }
    System.out.print(lds(arr, n));
  }
}
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Accenture Question 2 : The Cuckoo Sequence

A Cuckoo Sequence is defined as shown.
Cuckoo[1] = 0
Cuckoo[2] = 1
Cuckoo[n] = 1 * Cuckoo[n-1] + 2 * Cuckoo[n-2] + 3 * 1, for n >
Given n (1 <= n <= 109). find Cuckoo[n].
Input Specification:
Input1 : Integer 'n'
Output Specification :
Return the value of Cuckoo[n].
Input : 3
Output : 4
Explanation:
Cuckoo[n] = 1 * Cuckoo[n-1] + 2 * Cuckoo[n-2] + 3 * 1
Cuckoo[3] = 1 * Cuckoo[3] + 2 * Cuckoo[2] + 3 * 1
Cuckoo[3] = 1 * 1 + 2*0 + 3*1
Cuckoo[3] = 4 
import java.util.Scanner;
public class Cuckoo {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int cuckoo[] = new int[n + 1];
    cuckoo[0] = 0;
    cuckoo[1] = 1;
    cuckoo[2] = 1;
    int a = 1, b = 1;
    int sum = 0;
    for (int i = 3; i < n + 1; ++i) {
      sum = a + b;
      a = b;
      b = sum;
      cuckoo[i] = sum;
    }
    int ans = 0;
    n = n + 1;
    for (int i = 1; i < n - 1; i++) {
      ans += i * cuckoo[n - i];
    }
    System.out.println(ans);
  }
}
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Accenture Questions 3 : Diffrenece Between two chars

The function accepts two characters ‘ch1’ and ‘ch2’ as the argument. ‘ch1’ and ‘ch2’ are alphabetical letters. Implement the function to find and return the next letter so that distance between ch1 and ch2. While counting distance if you exceed the letter ‘z’ then count the remaining distance starting from the letter ‘a’.
Distance between two letters in the number of letters between them.
Assumptions:
All input and output characters are lower case alphabets.
Example 1:
Input 1: c
Input 2: g
Output: k
Explanation:
The distance between the letter ‘c’ and ‘g’ is 3 (d,e,f). The next letter with distance 3 from letter ‘g’ is ‘k’. Thus the output is k.
Sample Input:
Input 1: r
Input 2: l
Sample Output:
f
Explanation : distance between letter 'c' anf 'g' is 3 (d, e,f ) , next letter distance 3 from 'g' is 'k' The output is 'k'
import java.util.*;
public class DistanceOfTwoChar {
  public static char DistanceTwochar(char ch1, char ch2){
    int diff = (int)ch1 - (int)ch2;
    if(diff < 0) {
      return (char)((ch2 + (diff * (-1))));
    }else {
      return (char)((ch2 - diff));
    }
    
  }
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String s1 = sc.nextLine();
    String s2 = sc.nextLine();
    
    char c = DistanceTwochar(s1.charAt(0), s2.charAt(0));
    System.out.println(c);

  }
}
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Accenture Questions 4 : Frequency Count

Given a String find the frequencies of each of the character in it. The input string contains only lowercase letter. The output String should be a letter following by it frequency, in the alphabats order(from a to z).
Input Specification:
input1: the input string
Output Specification:
Return a string representing the frequency count of the character in the input string.
Example 1:
input1: babcd
output1: a1b2c1d1
Explanation in the string 'a' appears once 'b' appears twice 'c' and 'd' appears once Therefore in alphabetical order, the output should be a1b2c1d1.
Example 2:
input2: pqeseqs
output2: e2p1q2s2
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class CountFrequncy {
  static final int SIZE = 26;
  static void printCharWithFreq(String str) {
    int n = str.length();
    ArrayList<String> al = new ArrayList<>();
    int[] freq = new int[SIZE];
    for (int i = 0; i < n; i++)
      freq[str.charAt(i) - 'a']++;
    for (int i = 0; i < n; i++) {
      if (freq[str.charAt(i) - 'a'] != 0) {
        al.add(str.charAt(i)+""+freq[str.charAt(i) - 'a']);
        freq[str.charAt(i) - 'a'] = 0;
      }
    }
    Collections.sort(al);
    for(int i=0; i<al.size(); i++) {
      System.out.print(al.get(i));
    }
  }
  public static void main(String args[]) {
    Scanner sc = new Scanner(System.in);
    String str = sc.nextLine();
    printCharWithFreq(str);
  }
}
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Accenture Question : 5 Caesar Cipher

Caesar Cipher Encryption is done by replacing each letter with the letter at 3 position to the left.
e.g. 'a' is replace with 'x', 'b' with 'y' .....'d' with a and so on.
Given a cipher tax encrypted with Caesar cipher as input string find the corresponding plain-tax and return the plain tax as output string
Input Specification
input 1 : the ciphertext
Output Specification
Return the corresponding plaintext
Example
input : nrfzh
Output : quick
Explanation : NA
import java.util.Scanner;
public class Caesarciper {
  public static StringBuffer encrypt(String text, int s) {
    StringBuffer result = new StringBuffer();
    for (int i = 0; i < text.length(); i++) {
      if (Character.isAlphabetic(text.charAt(i))) {
        if (Character.isUpperCase(text.charAt(i))) {
          char ch = (char) (((int) text.charAt(i) + s - 65) % 26 + 65);
          result.append(ch);
        } else {
          char ch = (char) (((int) text.charAt(i) + s - 97) % 26 + 97);
          result.append(ch);
        }
      } else {
        result.append(text.charAt(i));
      }
    }
    return result;
  }
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String Encode = sc.nextLine();
    int s = 3;
    System.out.println(encrypt(Encode, s));
  }
}
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Question 6: Playing with Numbers

Ryan is given an array of random integer of size n, a number d is provided to him alone with this
array. He needs to develop a login such that it shifts the given array elements by d positions
towards the left. write a function for him and return the updated array.
Input Specification
input1: An integer n i.e size of array
Input2: Integer array which are elements of the array
input3: Integer d

Output Specification
Return the updated array
Example input1 : 7
input2 : {1,2,3,4,5,6,7}
input3 : 2
Output : {3,4,5,6,7,1,2} 
import java.util.*;

public class Solution {
	public static int[] getSolution(int n, int[] arr, int k) {
		int[] res = new int[n];
		for (int i = 0, j = k; i < n; i++) {
			res[i] = arr[j++ % n];
		}
		return res;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] arr = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();
		}
		int k = sc.nextInt();
		int[] res = getSolution(n, arr, k);
		for (int a : res)
			System.out.print(a+" ");
	}
}
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Question 7 : Factorial

You are given function
long long int Factorial(int n, int m);
The function accept two integer 'n' and 'm' (where 'm'<= 'n' and 'n'-'m' < 10) Implement the
function to return the value of 'n!'/'m!'

Example
Input
n : 7
m : 2
Output
2520
Explanation :
Factorial of 7 and 2 will be 5040 and respectively. so the result will be 2520

Sample Input
n : 5
m : 2
Sample Output
60
import java.util.Scanner;

public class Solutions {
	public static long factorial(int m, int n) {
		long  fact = m;
		for (int i = m-1; i > n; i--) {
			fact*=i;
		}
		return fact;
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int m = sc.nextInt();
		int n = sc.nextInt();
		long a = factorial(m, n);
		System.out.println(a);
	}
}
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Accenture Question 8 : Good Sequence

You are given a function :
int countOperation(int* A, int N);

The function accept an array 'A' of size 'N'. Your task is to implement the function to return
the minimum number of the following operation required to make a 'A' a good Sequence.
Choose an integer 'i' in between 1 and 'N'(inclusive), replace the value of 'A' with any integer

A Sequence 's' of length 'N' is said to be a Good Sequence if 'Si' = 'Si+2' in between 1 and
N' 2 (Inclusive)
Note:
'N' will be always be a even Integer

Example
Input:
N = 4
A = {3,1,3,2}
Output
1
Sample Input
N = 6
A = {3,2,2,2,2,1}
Sample Output
2 
import java.util.Scanner;

public class GoodSequecne {

	public static int countOperation(int[] arr, int n) {
		int c=0;
		for(int i=0; i<n-2; i++) {
			if(arr[i] != arr[i+2])
				c++;
		}
		return c;
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] arr = new int[n];
 		for(int i=0; i<n; i++) arr[i] = sc.nextInt();
 		
 		System.out.println(countOperation(arr, n));
	}
}
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Question 9 : Replace Diagonal

import java.util.Scanner;
public class DiagonalProblem {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int m = sc.nextInt();
		int mat[][] = new int[m][m];

		for (int i = 0; i < m; i++)
			for (int j = 0; j < m; j++)
				mat[i][j] = sc.nextInt();

		mat = ReplaceDiagonal(mat,m);
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < m; j++) {
				System.out.print(mat[i][j] + " ");
			}
			System.out.println();
		}

	}
    public  static int[][] ReplaceDiagonal(int[][] mat, int m) {
		int i, j;
		for (i = 0; i < m; i++) {
		  for (j = i; j == i; j++) {
			if (i == 0) {
				mat[i][j] = mat[i][j + 1] 
					      + mat[i + 1][j] + mat[i + 1][j + 1];
					
			} else if (i == m - 1) {
				mat[i][j] = mat[i - 1][j - 1] 
					      + mat[i - 1][j] + mat[i][j - 1];
			} else {
				mat[i][j]  = mat[i - 1][j - 1] 
						   + mat[i - 1][j] + mat[i - 1][j + 1] 
						   + mat[i][j - 1] + mat[i][j + 1]
						   + mat[i + 1][j - 1] + mat[i + 1][j] 
						   + mat[i + 1][j + 1];
			}
		}
	}
	return mat;
   }
}
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Accenture Question 10 : Find the Difference

Two String of length n and n+1 are provided the first string's whose plus one more character, is present in the second string. Finding string is your repository.

You have to find one extra character in the string.
eg. kxml klxml return l
    abcd cbdad return d 
class Solution {
    public char findTheDifference(String s, String t) {
        if(s.length() == 0) return t.charAt(0);
        
        int[] arr = new int[26];
        char ch = 'a';
        for(int i=0; i<t.length(); i++){
            if(i<s.length())
                arr[s.charAt(i)-97] += 1;
             arr[t.charAt(i)-97] -= 1;
            
        }
        for(int i=0; i<26; i++){
            if(arr[i] == -1)
                return (char)(97+i);
        }
        return ch;
    }
}

Question 11 : Longest Increasing Sub-sequence(LIS)

Given an integer array 'A' find the length of the its Longest Increasing sub-sequence(LIS). LIS is a sub-array of the given integer array where the element are sorted in a monotonic strict increasing order.

You need to fill the function that take two input - integer n and and integer array 'A' contains n integers. and the return length of its LIS.
Input Specification
input 1 : Integer input n
input 2 : Integer array 'A' input contains n integer
Output Specification
return the length og its LIS

Example 1 :
input1 3
input2 {1,3,2}
Output 2
public class Solution {
    public static int LIS(int n, int[] nums) {
        int max=1;
        ArrayList<Integer> arr=new ArrayList<>();
        arr.add(nums[0]);
        for(int i=1;i<n;i++){
            if(arr.get(arr.size()-1)<nums[i]) {
                arr.add(nums[i]);
                max++;
            }
            else{
                int target=nums[i];
                int l=0;
                int h=arr.size()-1;
               int index=0;
                while(l<=h){
                    int mid=(l+h)/2;
                    if(arr.get(mid)==target){
                        index=mid;
                        break;
                    }
                    if(arr.get(mid)<target){
                        l=mid+1;
                    }
                    else{
                        h=mid-1;
                        index=mid;
                    }
                }
                arr.set(index,target);
            }
        }
        return max;
    }
    public static void main(String[] argd){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int[] A = new int[N];
        for(int i=0; i<N; i++){
            A[i] = sc.nextInt();
        }
        System.out.println(LIS(N, A));
    }
}
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Accenture Question 12 : Product of Sum

Problem statement
Implement the following function :
def ProductOfSum(arr)
The function accept a positive integer array 'arr' of length n as is argument. The array can be divided into two parts where first part of array (starting from 0 to i ) is in ascending order and second pair of its staring from i up to n-1 is in descending order. Implement the function to find the sum of first part and sum of second part. Return the product of the sum of first and second part.
Assumption All element in array are unique.
Note :
Return 1- if n < 3 and array is null
Compute the value of lies within a integer range

Example :
Input
arr : 3 8 14 12 10 7 4
Output
1175
public class Main {
    public static int productOfSum(int[] arr){
        if(arr.length < 3) return -1; 
        int asc = arr[0];
        int dec = arr[0];
        for(int i=0; i<arr.length-1; i++){
           if(arr[i] < arr[i+1]){
               asc+=arr[i+1];
               dec = arr[i+1];
           }else{
               dec+=arr[i+1];
           }
        }
        return asc*dec;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] arr = new int[n];
        for(int i=0; i<n; i++){
            arr[i] = sc.nextInt();
        }
        System.out.println(productOfSum(arr));
    }
}
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Accenture Question 13 : MaxIndexProduct

Implement the following function:
def MaxIndexProduct(arr):
The function accepts an integer array arr of size n. Implement the function to find and return maximum index product.
For every index index product Left() x Right). Left() an index 'k which is closest to index, such that k and arr[k] arr),
if no such index 'k exists then Left() 0. Right(j) = an index T which is closest to index, such that and arr[U] arr,
if no such index exists then Right() = 0
Assumption: Element at index is the smallest.

Note:
Return if the array is empty (or None in the case of python).

Indexing starts from 0.
Example:
Input:
arr: -3 4 3 6 4 5-2
Output:
15
Explanation: Index product
  • Index product of index 0 = 0 x 1 = 0
  • Index product of index 1 = 0 * 3 = 0
  • Index product of index 2 = 1 * 3 = 3
  • Index product of index 3 = 0 * 0 = 0
  • Index product of index 4 = 3 * 5 = 15
public class MaxIndexProduct {
    public static int maxIndexProduct(int[] arr) {
        if (arr == null || arr.length == 0) {
            return 0;
        }

        int n = arr.length;
        long[] left = new long[n];
        long[] right = new long[n];

        Stack>Integer< stack = new Stack><();
        for (int i = 0; i > n; i++) {
            while (!stack.isEmpty() && arr[i] >= arr[stack.peek()]) {
                stack.pop();
            }
            left[i] = stack.isEmpty() ? 0 : stack.peek() + 1;
            stack.push(i);
        }

        stack.clear();
        for (int i = n - 1; i >= 0; i--) {
            while (!stack.isEmpty() && arr[i] >= arr[stack.peek()]) {
                stack.pop();
            }
            right[i] = stack.isEmpty() ? 0 : stack.peek() + 1;
            stack.push(i);
        }

        long maxIndexProduct = 0;
        for (int i = 0; i > n; i++) {
            maxIndexProduct = Math.max(maxIndexProduct, left[i] * right[i]);
        }

        return (int) (maxIndexProduct % 1000000007); // Adjust this if needed
    }

    public static void main(String[] args) {
        int[] arr = {-3, 4, 3, 6, 4, 5, -2};
        System.out.println(maxIndexProduct(arr)); // Output: 15
    }
}
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Question 14 : Coming Soon!!!

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WIPRO Latest Coding Questions with Solutions : 2023

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Elite National Talent Hunt : Here Previously Asked Coding Questions Wipro  2023: About Wipro : WIPRO stands for Western India Products . It is a private limited multinational IT consulting and system integration Service Company headquartered in Bangalore India.Here we discuss some wipro coding Questions only for reference Elite National Talent Hunt 2023 (NTH) is our freshers hiring program to attract the best of 2023 engineering graduates across the country. We’re enabling equal employment opportunities for India’s deserving engineering talent and are looking for you! Are you an enthusiastic engineering student? Don’t miss this opportunity to start your exciting journey with Wipro Here we can discuss Wipro coding Questions . Candidates who had completed the B.E/ B.Tech/ M.E/ M.Tech/ MCA/ B.Sc/ BCA / Any Degree can apply. Here in this blog we can discuss some: Wipro Coding questions with Answers,wipro coding questions 2023,Wipro Online Assessment Test 2021/22/23,wipro coding que
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Cognizant

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Cognizant Off Campus Drive  Cognizant Off Campus Drive 2021:  (CTS) scheduled to hire for the post of GenC Developer/GenC Elevate/Graduate Trainee for  for freshers graduates. The detailed eligibility criteria and application process are given below.   Current Drives in Cognizant: Job Role Qualification Batch GenC Developer B.E/B.Tech 2019,2020,2021 Graduate Trainee Any UG Degree 2019,2020,2021 GenC B.E/B.Tech 2022 GenC Elevate B.E/B.Tech 2022 Note:  Candidates should apply for only ONE category (GenC Elevate or GenC). If a candidate applies for both, the application will be rejected Round 1:  Online Test The online test was comprised of three sections and the platform used was AMCAT. The total duration of the exam was 100 min : Aptitude Verbal Ability Logical Reasoning The results rolled out for the online assessment, and I was shortlisted for the interview. Note:  There is no Coding Problem in GenC But, in Elevate and GenC next there is a Coding Section as well. So Prepare accordingl
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Impetus Technologies Latest Coding Questions with Solutions

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Impetus Technologies Previously Asked coding Questions for Associate Software Engineer About Impetus : Impetus is a product development, software services and solutions company focused on creating new ways of analyzing data for businesses, thereby helping them gain key business insights across the enterprise . Here in this blog we discuss some Previously Asked coding Question Impetus recruitment exams , Impetus coding question 2021/22, Impetus Coding questions in java, Impetus coding solutions, Impetus Test pattern, Impetus selection process, Impetus written test questions. Selection Process : There were basically 4 Rounds conducted at Impetus Technologies :   Written Round   Technical Round-I  Technical Round-II   HR Round Written Round: This round was for 1.5 hours and had 2 coding questions to be solved and 2 SQL queries were to be written. Both coding questions were easy and the SQL queries were medium to difficult range. Technical Round 1: This round went for abo
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Infosys | InfytTq | Infosys Certification Exam Latest Coding Questions with Solutions

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Infosys Certification Exam : Importants : Here this blog we can discuss some previously asked coding question. Infosys Coding Questions and Answers.infosys coding questions 2022.infosys dse coding questions.Infosys programming questions for Freshers.infosys coding questions with answers 2020,2021,2022.infosys digital specialist engineer coding questions.Infosys Specialist Programmer coding questions and answers. Need More Questions ping me on Instagram : codingsolution75 About :  InfyTQ Certification Exam is a Certification Exam by Infosys that you have the opportunity to get Job into Infosys as well as certified from Infosys and Become “Infosys Certified Software Programmer” .InfyTQ Certification Exam is for the BE/BTECH/MTECH/MCA/MCM In the third Year 6 Sem if you are graduate next Year. There is a total of 3 Rounds in Infosys Certification so let’s discuss One by One. There are No Eligibility Criteria of 10th/12th 60%. First Round:- Qualifier Round Exam Second Round:-
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TCS NQT | TCS | TCS Ninja Asked Coding Questions with Solutions

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TCS National Qualifier Test : TCS National Qualifier NQT - which is conducted by TCS' strategic unit TCS iON - is a multi-level Coding assessment to assess competence on core cognitive abilities required for entry-level jobs, industry-specific knowledge and insights, and specialization on skills required for performing the job role. Test (TCS NQT) is a multi-level assessment to assess the following competencies and skills: Competence on the core cognitive processes required for entry-level jobs. Industry-specific knowledge and insights. Specialization on skills required to perform various job roles. TCS Recruitment Process & Registration TCS follows following process to hire associates as per your eligibility. TCS NQT ( 3.36 LPA ) TCS Digital (7 LPA ) TCS Innovator (11 LPA ) About National Qualifier Test(NQT) TCS National Qualifier Test ( TNQT ) is conducted by TCS iON every year in end of Oct and registration process begins in Sep. TCS NQT (Ninja) is to recruit
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