LeetCode //C - 1300. Sum of Mutated Array Closest to Target

1300. Sum of Mutated Array Closest to Target

Given an integer array arr and a target value target, return the integer value such that when we change all the integers larger than value in the given array to be equal to value, the sum of the array gets as close as possible (in absolute difference) to target.

In case of a tie, return the minimum such integer.

Notice that the answer is not neccesarilly a number from arr.

Example 1:

Input: arr = 4,9,3, target = 10

Output: 3

Explanation: When using 3 arr converts to 3, 3, 3 which sums 9 and that's the optimal answer.

Example 2:

Input: arr = 2,3,5, target = 10

Output: 5

Example 3:

Input: arr = 60864,25176,27249,21296,20204, target = 56803

Output: 11361

Constraints:

• 1 < = a r r . l e n g t h < = 10 4 1 <= arr.length <= 10^4 1<=arr.length<=104
• 1 < = a r r i , t a r g e t < = 10 5 1 <= arri, target <= 10^5 1<=arri,target<=105

From: LeetCode

Link: 1300. Sum of Mutated Array Closest to Target

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Solution:

Ideas:

Binary search the smallest value whose mutated array sum is at least target, then compare it with the previous value.

Code:

#include <stdlib.h>

int findBestValue(int* arr, int arrSize, int target) {
    int left = 0;
    int right = 0;

    for (int i = 0; i < arrSize; i++) {
        if (arr[i] > right) {
            right = arr[i];
        }
    }

    while (left < right) {
        int mid = left + (right - left) / 2;
        int sum = 0;

        for (int i = 0; i < arrSize; i++) {
            sum += arr[i] < mid ? arr[i] : mid;
        }

        if (sum < target) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }

    int sum1 = 0;
    int sum2 = 0;

    for (int i = 0; i < arrSize; i++) {
        sum1 += arr[i] < left ? arr[i] : left;
        sum2 += arr[i] < left - 1 ? arr[i] : left - 1;
    }

    if (abs(sum2 - target) <= abs(sum1 - target)) {
        return left - 1;
    }

    return left;
}