Array Manipulation and Sum using C/C++

In this problem, you are given an integer array arr of size n and an integer S. Your task is to find an element k in the array such that if all the elements greater than k in the array are replaced with k, then the sum of all the elements of the resultant array becomes equal to S. If such an element exists, print it; otherwise, print -1.

Scenario 1:

Input:
    int S = 9;
    int arr[] = { 1, 3, 2, 5, 8 };

Output:
    2

Explanation: In the given array, if we replace all elements greater than 2 with 2, the resultant array becomes {1, 2, 2, 2, 2}. The sum of this array is 9, which is equal to S. So the output is 2.

Scenario 2:

Input:
    int S = 12;
    int arr[] = { 1, 3, 2, 5, 8 };

Output:
    3

Explanation: In the given array, if we replace all elements greater than 3 with 3, the resultant array becomes {1, 3, 2, 3, 3}. The sum of this array is 12, which is equal to S. So the output is 3.

Algorithm to Solve Array Manipulation and Sum

To solve this problem, we can follow these steps:

• Initialize an array arr of size n and an integer S.
• Sort the array in ascending order.
• Initialize a variable sum to 0.
• Traverse through the sorted array and for each element, check if sum + (arr[i] * (n - i)) = S, where sum consists of the sum of all elements before the current index and i is the current index of iteration.
• If the condition is true, print K and end the program.
• If not, add the current element to sum and continue the next iteration of the loop.
• If you are able to traverse the entire array without finding such an element, print -1.

C/C++ Code to Implement Array Manipulation and Sum

Here is a C/C++ implementation of the algorithm described above:

#include <stdio.h>
#include <stdlib.h>

// comparator function for qsort
int cmpfunc(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int getElement(int arr[], int n, int S) {
    // Sort the array
    qsort(arr, n, sizeof(int), cmpfunc);

    int sum = 0;

    for (int i = 0; i < n; i++) {
        // If current element satisfies the condition
        if (sum + (arr[i] * (n - i)) == S)
            return arr[i];
        sum += arr[i];
    }

    // No element found
    return -1;
}

int main() {
    int S = 9;
    int arr[] = { 1, 3, 2, 5, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);

    printf("%d\n", getElement(arr, n, S));

    return 0;
}

2

#include <bits/stdc++.h>
using namespace std;

// Function to return the required element
int getElement(int a[], int n, int S)
{
    // Sort the array
    sort(a, a + n);
    int sum = 0;

    for (int i = 0; i < n; i++) {

        // If current element
        // satisfies the condition
        if (sum + (a[i] * (n - i)) == S)
            return a[i];
        sum += a[i];
    }

    // No element found
    return -1;
}

// Driver code
int main()
{
    int S = 12;
    int a[] = { 1, 3, 2, 5, 8 };
    int n = sizeof(a) / sizeof(a[0]);

    cout << getElement(a, n, S);

    return 0;
}

3