Welcome to Subscribe On Youtube

1300. Sum of Mutated Array Closest to Target

Description

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 <= arr.length <= 104
• 1 <= arr[i], target <= 105

Solutions

• Java
• C++
• Python
• Go

• class Solution {
      public int findBestValue(int[] arr, int target) {
          Arrays.sort(arr);
          int n = arr.length;
          int[] s = new int[n + 1];
          int mx = 0;
          for (int i = 0; i < n; ++i) {
              s[i + 1] = s[i] + arr[i];
              mx = Math.max(mx, arr[i]);
          }
          int ans = 0, diff = 1 << 30;
          for (int value = 0; value <= mx; ++value) {
              int i = search(arr, value);
              int d = Math.abs(s[i] + (n - i) * value - target);
              if (diff > d) {
                  diff = d;
                  ans = value;
              }
          }
          return ans;
      }
  
      private int search(int[] arr, int x) {
          int left = 0, right = arr.length;
          while (left < right) {
              int mid = (left + right) >> 1;
              if (arr[mid] > x) {
                  right = mid;
              } else {
                  left = mid + 1;
              }
          }
          return left;
      }
  }
  

• class Solution {
  public:
      int findBestValue(vector<int>& arr, int target) {
          sort(arr.begin(), arr.end());
          int n = arr.size();
          int s[n + 1];
          s[0] = 0;
          int mx = 0;
          for (int i = 0; i < n; ++i) {
              s[i + 1] = s[i] + arr[i];
              mx = max(mx, arr[i]);
          }
          int ans = 0, diff = 1 << 30;
          for (int value = 0; value <= mx; ++value) {
              int i = upper_bound(arr.begin(), arr.end(), value) - arr.begin();
              int d = abs(s[i] + (n - i) * value - target);
              if (diff > d) {
                  diff = d;
                  ans = value;
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def findBestValue(self, arr: List[int], target: int) -> int:
          arr.sort()
          s = list(accumulate(arr, initial=0))
          ans, diff = 0, inf
          for value in range(max(arr) + 1):
              i = bisect_right(arr, value)
              d = abs(s[i] + (len(arr) - i) * value - target)
              if diff > d:
                  diff = d
                  ans = value
          return ans
  
  

• func findBestValue(arr []int, target int) (ans int) {
  	sort.Ints(arr)
  	n := len(arr)
  	s := make([]int, n+1)
  	mx := slices.Max(arr)
  	for i, x := range arr {
  		s[i+1] = s[i] + x
  	}
  	diff := 1 << 30
  	for value := 0; value <= mx; value++ {
  		i := sort.SearchInts(arr, value+1)
  		d := abs(s[i] + (n-i)*value - target)
  		if diff > d {
  			diff = d
  			ans = value
  		}
  	}
  	return
  }
  
  func abs(x int) int {
  	if x < 0 {
  		return -x
  	}
  	return x
  }
  

All Problems

All Solutions