1551. Minimum Operations to Make Array Equal LeetCode Solution

Last updated: February 24, 2025

In this guide, you will get 1551. Minimum Operations to Make Array Equal LeetCode Solution with the best time and space complexity. The solution to Minimum Operations to Make Array Equal problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Minimum Operations to Make Array Equal solution in C++
Minimum Operations to Make Array Equal solution in Java
Minimum Operations to Make Array Equal solution in Python
Additional Resources

1551. Minimum Operations to Make Array Equal LeetCode Solution image

Problem Statement of Minimum Operations to Make Array Equal

You have an array arr of length n where arr[i] = (2 * i) + 1 for all valid values of i (i.e., 0 <= i < n).
In one operation, you can select two indices x and y where 0 <= x, y < n and subtract 1 from arr[x] and add 1 to arr[y] (i.e., perform arr[x] -=1 and arr[y] += 1). The goal is to make all the elements of the array equal. It is guaranteed that all the elements of the array can be made equal using some operations.
Given an integer n, the length of the array, return the minimum number of operations needed to make all the elements of arr equal.

Example 1:

Input: n = 3
Output: 2
Explanation: arr = [1, 3, 5]
First operation choose x = 2 and y = 0, this leads arr to be [2, 3, 4]
In the second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].

Example 2:

Input: n = 6
Output: 9

Constraints:

1 <= n <= 104

Complexity Analysis

Time Complexity: O(1)
Space Complexity: O(1)

1551. Minimum Operations to Make Array Equal LeetCode Solution in C++

class Solution {
 public:
  int minOperations(int n) {
    //     median := median of arr
    //   diffs[i] := median - arr[i] where i <= i <= n / 2
    //        ans := sum(diffs)
    // e.g.
    // n = 5, arr = [1, 3, 5, 7, 9], diffs = [4, 2]
    //        ans = (4 + 2) * 2 / 2 = 6
    // n = 6, arr = [1, 3, 5, 7, 9, 11], diffs = [5, 3, 1]
    //        ans = (5 + 1) * 3 / 2 = 9
    const int halfSize = n / 2;
    const int median = (arr(n) + arr(1)) / 2;
    const int firstDiff = median - arr(1);
    const int lastDiff = median - arr(halfSize);
    return (firstDiff + lastDiff) * halfSize / 2;
  }

 private:
  // Returns the i-th element of `arr`, where 1 <= i <= n.
  int arr(int i) {
    return (i - 1) * 2 + 1;
  }
};
/* code provided by PROGIEZ */

1551. Minimum Operations to Make Array Equal LeetCode Solution in Java

class Solution {
  public int minOperations(int n) {
    //     median := median of arr
    //   diffs[i] := median - arr[i] where i <= i <= n / 2
    //        ans := sum(diffs)
    // e.g.
    // n = 5, arr = [1, 3, 5, 7, 9], diffs = [4, 2]
    //        ans = (4 + 2) * 2 / 2 = 6
    // n = 6, arr = [1, 3, 5, 7, 9, 11], diffs = [5, 3, 1]
    //        ans = (5 + 1) * 3 / 2 = 9
    final int halfSize = n / 2;
    final int median = (arr(n) + arr(1)) / 2;
    final int firstDiff = median - arr(1);
    final int lastDiff = median - arr(halfSize);
    return (firstDiff + lastDiff) * halfSize / 2;
  }

  // Returns the i-th element of `arr`, where 1 <= i <= n.
  private int arr(int i) {
    return (i - 1) * 2 + 1;
  }
}
// code provided by PROGIEZ

1551. Minimum Operations to Make Array Equal LeetCode Solution in Python

class Solution:
  def minOperations(self, n: int) -> int:
    def arr(self, i: int) -> int:
      """Returns the i-th element of `arr`, where 1 <= i <= n."""
      return (i - 1) * 2 + 1

    #     median := median of arr
    #   diffs[i] := median - arr[i] where i <= i <= n // 2
    #        ans := sum(diffs)
    # e.g.
    # n = 5, arr = [1, 3, 5, 7, 9], diffs = [4, 2]
    #        ans = (4 + 2) * 2 // 2 = 6
    # n = 6, arr = [1, 3, 5, 7, 9, 11], diffs = [5, 3, 1]
    #        ans = (5 + 1) * 3 // 2 = 9
    halfSize = n // 2
    median = (arr(n) + arr(1)) // 2
    firstDiff = median - arr(1)
    lastDiff = median - arr(halfSize)
    return (firstDiff + lastDiff) * halfSize // 2
 # code by PROGIEZ

Additional Resources

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