Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Move from brute-force thinking to an efficient approach using math strategy.
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 <= 104Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math
3
6
minimum-number-of-operations-to-make-arrays-similar)minimum-operations-to-make-array-equal-ii)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1551: Minimum Operations to Make Array Equal
class Solution {
public int minOperations(int n) {
int ans = 0;
for (int i = 0; i < n >> 1; ++i) {
ans += n - (i << 1 | 1);
}
return ans;
}
}
// Accepted solution for LeetCode #1551: Minimum Operations to Make Array Equal
func minOperations(n int) (ans int) {
for i := 0; i < n>>1; i++ {
ans += n - (i<<1 | 1)
}
return
}
# Accepted solution for LeetCode #1551: Minimum Operations to Make Array Equal
class Solution:
def minOperations(self, n: int) -> int:
return sum(n - (i << 1 | 1) for i in range(n >> 1))
// Accepted solution for LeetCode #1551: Minimum Operations to Make Array Equal
struct Solution;
impl Solution {
fn min_operations(n: i32) -> i32 {
let mut res = 0;
let mut i = 1;
while i < n {
res += n - i;
i += 2;
}
res
}
}
#[test]
fn test() {
let n = 3;
let res = 2;
assert_eq!(Solution::min_operations(n), res);
let n = 6;
let res = 9;
assert_eq!(Solution::min_operations(n), res);
}
// Accepted solution for LeetCode #1551: Minimum Operations to Make Array Equal
function minOperations(n: number): number {
let ans = 0;
for (let i = 0; i < n >> 1; ++i) {
ans += n - ((i << 1) | 1);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.
Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.