**Accenture Hack Diva Demo Round Coding Question 2024 | Accenture Hack Diva Demo Round Coding Question & Answers | Accenture Hack Diva Demo Round Coding Question With Solutions**

**Accenture Hack Diva Demo Round Coding Question 2024: **The Accenture Hack Diva Demo Round coding test is medium complexity, meaning that while passing it is not impossible, it is also not very simple. If you practiced some of the code questions from the previous year’s examination, you would undoubtedly have an advantage in the upcoming Accenture Hack Diva Demo Round coding exams. Let’s look at a few test patterns, Accenture Hack Diva Demo Round Coding Questions and Answers/ Accenture Hack Diva Demo Round Coding Questions in Python, C, C++, Java, and Python code problems and their fixes. In the below section, you can find the Accenture Hack Diva Demo Round Coding Exam Questions and Answers/Accenture Hack Diva Demo Round **Coding Questions** & Answers 2024

**Accenture Hack Diva Demo Round Coding Exam Solved Questions**

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**Accenture Hack Diva Demo Round Coding Question **

**Q)** **Write a program to calculate and return the sum of absolute difference between the adjacent number in an array of positive integers from the position entered by the user.**

**Note** : You are expected to write code in the **findTotalSum **function only which receive three positional arguments:

1st : number of elements in the array

2nd : array

3rd : position from where the sum is to be calculated

**Example**

**Input
**input 1 : 7

input 2 : 11 22 12 24 13 26 14

input 3 : 5

**Output
**25

**Explanation**

The first parameter 7 is the size of the array. Next is an array of integers and input 5 is the position from where you have to calculate the Total Sum. The output is 25 as per calculation below.

| 26-13 | = 13

| 14-26 | = 12

Total Sum = 13 + 12 = 25

**Here is the code:**

#include <stdio.h> #include <stdlib.h> int findTotalSum(int n, int arr[], int start) { int difference, sum=0; for(int i=start-1; i<n-1; i++) { difference = abs(arr[i]-arr[i+1]); sum = sum + difference; } return sum; } int main() { int n; int start; scanf("%d",&n); int array[n]; for(int i=0; i<n; i++) { scanf("%d",&array[i]); } scanf("%d",&start); int result = findTotalSum(n, array, start); printf("\n%d",result); return 0; }

**Q)** **Write a program to return the difference between the count of odd numbers and even numbers**.

You are expected to write code in the **countOddEvenDifference **function only which will receive the first parameter as the number of items in the array and second parameter as the array itself. you are not required to take input from the console.

**Example
**Finding the difference between the count of odd and even numbers from a list of 5 number

**Input
**input 1 : 8

input 2 : 10 20 30 40 55 66 77 83

**Output
**-2

**Explanation
**The first paramter (8) is the szie of the array. Next is an array of integers. The calculation of difference between count sum of odd and even numbers is as follows:

**3 (count of odd numbers) – 5 (count of even numbers) = -2**

**Here is the code:**

#include <stdio.h> int countOddEvenDifference(int n, int arr[]) { int odd = 0, even = 0; for(int i=0; i<n; i++) { if(arr[i]%2==0) { even = even+1; } else { odd = odd+1; } } return odd - even; } int main() { int n; scanf("%d",&n); int array[n]; for(int i=0; i<n; i++) { scanf("%d",&array[i]); } int result = countOddEvenDifference(n, array); printf("%d",result); return 0; }

**Question) **Rohan and his team are participating in the Treasure Hunt event of college in which in each step they have to solve one problem to get a clue about the Treasure location. Rohan’s team has performed very well and reached the final step where they have to provide a code of a problem to get a final clue about treasure .

Given a string s, they need to find the longest palindromic subsequence’s length in s.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

The string contains only lowercase letters.

Write a program to help Rohan’s team that takes in input as String x and returns the length of the longest palindromic subsequence of x.

**Input Specification:**

**input1**: string input

**Output Specification:**

Return the length of the longest palindromic subsequence

**Example 1:**

**Input:** s = “bbbab”

**Output:** 4

**Explanation:** One possible longest palindromic subsequence is “bbbb”.

**Example 2:**

**Input:** s = “cbbd”

**Output:** 2

**Explanation:** One possible longest palindromic subsequence is “bb”.

**C++ PROGRAM**

#include<iostream> #include<cstring> using namespace std; int max (int x, int y) { return (x > y)? x : y; } int lps(char *str) { int n = strlen(str); int i, j, cl; int L[n][n]; for (i = 0; i < n; i++) L[i][i] = 1; for (cl = 2; cl <= n; cl++) { for (i = 0; i < n-cl+1; i++) { j = i + cl-1; if (str[i] == str[j] && cl == 2) L[i][j] = 2; else if (str[i] == str[j]) L[i][j] = L[i+1][j-1] + 2; else L[i][j] = max(L[i][j-1], L[i+1][j]); } } return L[0][n-1]; } int main() { char seq[20]; cin >> seq; int n = strlen(seq); printf("%d", lps(seq)); getchar(); return 0; }

**JAVA PROGRAM**

import java.util.*; public class Main { static int max (int x, int y) { return (x > y)? x : y; } static int lps(String seq) { int n = seq.length(); int i, j, cl; int L[][] = new int[n][n]; for (i = 0; i < n; i++) L[i][i] = 1; for (cl = 2; cl <= n; cl++) { for (i = 0; i < n-cl+1; i++) { j = i+cl-1; if (seq.charAt(i) == seq.charAt(j) && cl == 2) L[i][j] = 2; else if (seq.charAt(i) == seq.charAt(j)) L[i][j] = L[i+1][j-1] + 2; else L[i][j] = max(L[i][j-1], L[i+1][j]); } } return L[0][n-1]; } public static void main(String args[]) { Scanner sc = new Scanner(System.in); String seq = sc.next(); int n = seq.length(); System.out.println(""+ lps(seq)); } }

**PYTHON PROGRAM**

s = input() n = len(s) rev = s[::-1] dp = [[0 for _ in range(n+1)] for _ in range(n+1)] for i in range(1, n+1): for j in range(1, n+1): if s[i-1] == rev[j-1]: dp[i][j] = 1 + dp[i-1][j-1] else: dp[i][j] = max(dp[i-1][j], dp[i][j-1]) print(dp[n][n])

**Questions:** A chef has n orders lined up. The waiting time penalty before an order starts getting processed is k. And for each order, a corresponding processing time t is given (in minutes) along with its order id. The earning per order is c times the time taken to process the order. Find out the profits earned for each order from 1 to n and print them in order of the input. The order is free if the profits are negative for that order i.e. the profits are max(0, profit).

The following rules are used for processing the orders:

- Pick the order which arrives first. The chef can process only one order at a time.
- Among the orders already arrived, pick the one with the least processing time.

**Input Format:**

First the number of test cases T is given. Then T test cases follow.

Each test case consists of n (the number of orders),c (the cost per minute it takes to process any order, which is the same for all orders), and k (the waiting time penalty per minute, the chef faces once an order has arrived and is there in the “waiting to be processed queue”).

Then in the following n lines, an order with its arrival time and the processing time is given.

The order id of the first of these n lines is 1, the order id of the 2nd order is 2, and so on till the nth line has the order id n.

**NOTE: **The input may or may not be sorted by the arrival time i.e. order 3 may arrive earlier than order 2 and so on.

**Output Format:** For each test case, print in a single line the profits earned for orders 1 to n, respectively.

**Constraints:**

1<=T<=10 ; 1<=n<=10^5 ; 1<=order arrival time, order processing time <= 10^9 ; 1<=k<=100 ; 1<=c<=10^4 ; Time limit: 1 second per test case

**Example :**

Input:

```
1
3 10 16
1 2
2 3
3 1
```

**Explanation: **

The number of test cases T = 1. A number of orders n = 3. Processing cost per order per minute i.e. c = 10. Penalty cost for an order already arrived but in the waiting queue per minute of waiting i.e. k = 1. Then the list of orders follows. Order 1 arrives at time t=1. the chef does nothing from t=0 to t=1.

Since he has to pick up the order which arrived first his profit earned for order 1 would be processing time*c — k*waiting time. Since the order was processed immediately as it came, the waiting time would be 0 for order 1. Hence, profit of order 1 = 2*10 — 16*0= 20. Now the time t=1+2=3. Therefore both orders 2 and 3 have arrived and are in the waiting queue.

But the chef chooses order 3 to be processed first since it has less processing time. Therefore profit for order 3 is 1*10-16*0=10, as order 3 was processed the moment it came. Now the time t=3+1=4 and the last order 2 is taken up which has been waiting for 4-2=2 minutes. Therefore the profit earned for order 2 = 3*10 — 16*2= 30 — 32 = -2. Therefore the order is free or profit = 0.

Output: 20 0 10

**Q2:**

**Solution:**

**#include**<bits/stdc++.h>

**using** **namespace** std;

**typedef** **long** **long** **int** ll;

**typedef** vector**<ll>** vi;

**#define** fo(i,s,e_ex) **for**(i=s;i<e_ex;i++)

**#define** pbb push_back

**#define** all(x) x.**begin**(),x.**end**()

**struct** task{

ll idx,at,bt;

};

**bool** comp(task p,task q){

**if**(p.at==q.at){

**return** p.bt<q.bt;

}

**return** p.at<q.at;

}

**bool** **Compare**(task a, task b){

**if**(a.bt==b.bt){

**return** a.idx>b.idx;

}

**return** a.bt>b.bt;

}

**void** solve(ll caseno){

ll i,j,n,k,d,t=0;

cin>>n>>k>>d;

vector**<task>** arr;

vector**<ll>** ans(n);

fo(i,0,n){

ll a,b;

cin>>a>>b;

task t;

t.idx=i;t.at=a;t.bt=b;

arr.pbb(t);

}

sort(all(arr),comp);

priority_queue<task, std::vector**<task>**, **decltype**(&**Compare**)> pq(**Compare**);

i=0;

**while**((!pq.empty()) || i<n){

**if**(pq.empty()){

pq.push(arr[i]);

t=arr[i].at;

i++;

}

task curr = pq.top();pq.pop();

ll wait = t-curr.at;

t+=curr.bt;

ll cost = curr.bt*k – wait*d;

cost = max(cost,0ll);

ans[curr.idx] = cost;

**while**(i<n && arr[i].at<=t){

pq.push(arr[i]);

i++;

}

}

**for**(**auto** v:ans) cout<<v<<**‘ ‘**;

cout<<endl;

}

**int** main(){

ios_base::sync_with_stdio(**false**);

cin.tie(0);cout.tie(0);

ll t=1;

cin>>t;

**for**(ll i=1;i<=t;i++){

solve(i);

}

**return** 0;

}

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