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.

Syntax

int getElement(int arr[], int n, int S);

Algorithm

To solve this problem efficiently, we follow these steps −

• Sort the array in ascending order.
• Initialize a variable sum to 0.
• For each element at index i, check if sum + (arr[i] * (n - i)) = S.
• If condition is true, return arr[i] as our answer.
• Otherwise, add current element to sum and continue.
• If no element satisfies the condition, return -1.

Example 1

Let's find an element k where replacing all elements greater than k makes the sum equal to 9 −

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

int cmpfunc(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int getElement(int arr[], int n, int S) {
    qsort(arr, n, sizeof(int), cmpfunc);
    
    int sum = 0;
    
    for (int i = 0; i < n; i++) {
        if (sum + (arr[i] * (n - i)) == S)
            return arr[i];
        sum += arr[i];
    }
    
    return -1;
}

int main() {
    int arr[] = {1, 3, 2, 5, 8};
    int n = sizeof(arr) / sizeof(arr[0]);
    int S = 9;
    
    printf("Target sum: %d\n", S);
    printf("Element k: %d\n", getElement(arr, n, S));
    
    return 0;
}

Target sum: 9
Element k: 2

Explanation: When k=2, we replace all elements greater than 2 with 2, giving us {1, 2, 2, 2, 2}. Sum = 1 + 2 + 2 + 2 + 2 = 9.

Example 2

Let's find k for a different target sum −

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

int cmpfunc(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int getElement(int arr[], int n, int S) {
    qsort(arr, n, sizeof(int), cmpfunc);
    
    int sum = 0;
    
    for (int i = 0; i < n; i++) {
        if (sum + (arr[i] * (n - i)) == S)
            return arr[i];
        sum += arr[i];
    }
    
    return -1;
}

int main() {
    int arr[] = {1, 3, 2, 5, 8};
    int n = sizeof(arr) / sizeof(arr[0]);
    int S = 12;
    
    printf("Target sum: %d\n", S);
    printf("Element k: %d\n", getElement(arr, n, S));
    
    return 0;
}

Target sum: 12
Element k: 3

Explanation: When k=3, we replace all elements greater than 3 with 3, giving us {1, 3, 2, 3, 3}. Sum = 1 + 3 + 2 + 3 + 3 = 12.

Time and Space Complexity

Conclusion

This array manipulation problem is efficiently solved by sorting the array first and then checking each element as a potential threshold value. The key insight is that we only need to check array elements as candidates for k.