Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
You are given an integer array nums and a non-negative integer k. In one operation, you can increase or decrease any element by 1.
Return the minimum number of operations needed to make the median of nums equal to k.
The median of an array is defined as the middle element of the array when it is sorted in non-decreasing order. If there are two choices for a median, the larger of the two values is taken.
Example 1:
Input: nums = [2,5,6,8,5], k = 4
Output: 2
Explanation:
We can subtract one from nums[1] and nums[4] to obtain [2, 4, 6, 8, 4]. The median of the resulting array is equal to k.
Example 2:
Input: nums = [2,5,6,8,5], k = 7
Output: 3
Explanation:
We can add one to nums[1] twice and add one to nums[2] once to obtain [2, 7, 7, 8, 5].
Example 3:
Input: nums = [1,2,3,4,5,6], k = 4
Output: 0
Explanation:
The median of the array is already equal to k.
Constraints:
1 <= nums.length <= 2 * 1051 <= nums[i] <= 1091 <= k <= 109Problem summary: You are given an integer array nums and a non-negative integer k. In one operation, you can increase or decrease any element by 1. Return the minimum number of operations needed to make the median of nums equal to k. The median of an array is defined as the middle element of the array when it is sorted in non-decreasing order. If there are two choices for a median, the larger of the two values is taken.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Greedy
[2,5,6,8,5] 4
[2,5,6,8,5] 7
[1,2,3,4,5,6] 4
find-median-from-data-stream)sliding-window-median)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3107: Minimum Operations to Make Median of Array Equal to K
class Solution {
public long minOperationsToMakeMedianK(int[] nums, int k) {
Arrays.sort(nums);
int n = nums.length;
int m = n >> 1;
long ans = Math.abs(nums[m] - k);
if (nums[m] > k) {
for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
ans += nums[i] - k;
}
} else {
for (int i = m + 1; i < n && nums[i] < k; ++i) {
ans += k - nums[i];
}
}
return ans;
}
}
// Accepted solution for LeetCode #3107: Minimum Operations to Make Median of Array Equal to K
func minOperationsToMakeMedianK(nums []int, k int) (ans int64) {
sort.Ints(nums)
n := len(nums)
m := n >> 1
ans = int64(abs(nums[m] - k))
if nums[m] > k {
for i := m - 1; i >= 0 && nums[i] > k; i-- {
ans += int64(nums[i] - k)
}
} else {
for i := m + 1; i < n && nums[i] < k; i++ {
ans += int64(k - nums[i])
}
}
return
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
# Accepted solution for LeetCode #3107: Minimum Operations to Make Median of Array Equal to K
class Solution:
def minOperationsToMakeMedianK(self, nums: List[int], k: int) -> int:
nums.sort()
n = len(nums)
m = n >> 1
ans = abs(nums[m] - k)
if nums[m] > k:
for i in range(m - 1, -1, -1):
if nums[i] <= k:
break
ans += nums[i] - k
else:
for i in range(m + 1, n):
if nums[i] >= k:
break
ans += k - nums[i]
return ans
// Accepted solution for LeetCode #3107: Minimum Operations to Make Median of Array Equal to K
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #3107: Minimum Operations to Make Median of Array Equal to K
// class Solution {
// public long minOperationsToMakeMedianK(int[] nums, int k) {
// Arrays.sort(nums);
// int n = nums.length;
// int m = n >> 1;
// long ans = Math.abs(nums[m] - k);
// if (nums[m] > k) {
// for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
// ans += nums[i] - k;
// }
// } else {
// for (int i = m + 1; i < n && nums[i] < k; ++i) {
// ans += k - nums[i];
// }
// }
// return ans;
// }
// }
// Accepted solution for LeetCode #3107: Minimum Operations to Make Median of Array Equal to K
function minOperationsToMakeMedianK(nums: number[], k: number): number {
nums.sort((a, b) => a - b);
const n = nums.length;
const m = n >> 1;
let ans = Math.abs(nums[m] - k);
if (nums[m] > k) {
for (let i = m - 1; i >= 0 && nums[i] > k; --i) {
ans += nums[i] - k;
}
} else {
for (let i = m + 1; i < n && nums[i] < k; ++i) {
ans += k - nums[i];
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.
Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.