Minimum Operations to Make Array Equal - Practice Coding Problems

Minimum Operations to Make Array Equal - Problem

Math Medium

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.

Input & Output

Example 1 — Even Length Array

$ Input: n = 3

› Output: 2

💡 Note: Array is [1,3,5]. Target is 3. Element 1 needs +2, element 5 has excess 2. Operations: 2.

Example 2 — Larger Even Array

$ Input: n = 6

› Output: 9

💡 Note: Array is [1,3,5,7,9,11]. Target is 6. Excess from second half: (7-6)+(9-6)+(11-6) = 1+3+5 = 9.

Example 3 — Single Element

$ Input: n = 1

› Output: 0

💡 Note: Array is [1]. Only one element, already equal. No operations needed.

Constraints

1 ≤ n ≤ 10⁴

Visualization

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Understanding the Visualization

Input Array

Array [1,3,5,7,9,11] with n=6

Target Value

All elements should become 6 (middle value)

Operations Needed

Count moves from excess to deficit: 9 operations

Key Takeaway

🎯 Key Insight: The target is always the middle value n, and we only need to sum the excess from the upper half

Asked in

G Google 15 M Microsoft 12

The key insight is that for array [1,3,5,7,...,2n-1], the target is always n. We only need to calculate excess from the second half using the mathematical formula. Best approach is the mathematical formula: Time: O(1), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Formula	O(1)	O(1)	Use mathematical insight to calculate operations directly without building array
Simulation Approach	O(n)	O(n)	Build the array and simulate operations until all elements are equal

Mathematical Formula — Algorithm Steps

Recognize target = n (middle element)
Only consider second half: elements > n
Sum formula: (n+1) + (n+3) + ... + (2n-1)
Apply arithmetic series formula

Visualization

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Step-by-Step Walkthrough

Pattern Recognition

Array [1,3,5,7,9,11], target = 6

Excess Calculation

Second half excess: [1,3,5] = 9 total

Formula

For even n: (n/2)² = (6/2)² = 9

Code -

solution.c — C

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

int solution(int n) {
    // For array [1,3,5,7,...,2n-1], target is n
    // Mathematical formula for minimum operations
    
    if (n % 2 == 0) {
        // Even n: operations = (n/2)²
        int half = n / 2;
        return half * half;
    } else {
        // Odd n: different pattern
        int k = (n - 1) / 2;
        return k * (k + 1);
    }
}

int main() {
    int n;
    scanf("%d", &n);
    
    int result = solution(n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Single arithmetic calculation using mathematical formula

✓ Linear Growth

Space Complexity

O(1)

Only uses constant extra variables for calculation

✓ Linear Space

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Input & Output

Constraints

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Common Approaches

Mathematical Formula — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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