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 a 0-indexed integer array nums and an integer p. Find p pairs of indices of nums such that the maximum difference amongst all the pairs is minimized. Also, ensure no index appears more than once amongst the p pairs.
Note that for a pair of elements at the index i and j, the difference of this pair is |nums[i] - nums[j]|, where |x| represents the absolute value of x.
Return the minimum maximum difference among all p pairs. We define the maximum of an empty set to be zero.
Example 1:
Input: nums = [10,1,2,7,1,3], p = 2 Output: 1 Explanation: The first pair is formed from the indices 1 and 4, and the second pair is formed from the indices 2 and 5. The maximum difference is max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1. Therefore, we return 1.
Example 2:
Input: nums = [4,2,1,2], p = 1 Output: 0 Explanation: Let the indices 1 and 3 form a pair. The difference of that pair is |2 - 2| = 0, which is the minimum we can attain.
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1090 <= p <= (nums.length)/2Problem summary: You are given a 0-indexed integer array nums and an integer p. Find p pairs of indices of nums such that the maximum difference amongst all the pairs is minimized. Also, ensure no index appears more than once amongst the p pairs. Note that for a pair of elements at the index i and j, the difference of this pair is |nums[i] - nums[j]|, where |x| represents the absolute value of x. Return the minimum maximum difference among all p pairs. We define the maximum of an empty set to be zero.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Binary Search · Dynamic Programming · Greedy
[10,1,2,7,1,3] 2
[4,2,1,2] 1
minimum-absolute-difference)minimum-difference-between-largest-and-smallest-value-in-three-moves)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2616: Minimize the Maximum Difference of Pairs
class Solution {
public int minimizeMax(int[] nums, int p) {
Arrays.sort(nums);
int n = nums.length;
int l = 0, r = nums[n - 1] - nums[0] + 1;
while (l < r) {
int mid = (l + r) >>> 1;
if (count(nums, mid) >= p) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
private int count(int[] nums, int diff) {
int cnt = 0;
for (int i = 0; i < nums.length - 1; ++i) {
if (nums[i + 1] - nums[i] <= diff) {
++cnt;
++i;
}
}
return cnt;
}
}
// Accepted solution for LeetCode #2616: Minimize the Maximum Difference of Pairs
func minimizeMax(nums []int, p int) int {
sort.Ints(nums)
n := len(nums)
r := nums[n-1] - nums[0] + 1
return sort.Search(r, func(diff int) bool {
cnt := 0
for i := 0; i < n-1; i++ {
if nums[i+1]-nums[i] <= diff {
cnt++
i++
}
}
return cnt >= p
})
}
# Accepted solution for LeetCode #2616: Minimize the Maximum Difference of Pairs
class Solution:
def minimizeMax(self, nums: List[int], p: int) -> int:
def check(diff: int) -> bool:
cnt = i = 0
while i < len(nums) - 1:
if nums[i + 1] - nums[i] <= diff:
cnt += 1
i += 2
else:
i += 1
return cnt >= p
nums.sort()
return bisect_left(range(nums[-1] - nums[0] + 1), True, key=check)
// Accepted solution for LeetCode #2616: Minimize the Maximum Difference of Pairs
impl Solution {
pub fn minimize_max(mut nums: Vec<i32>, p: i32) -> i32 {
nums.sort();
let n = nums.len();
let (mut l, mut r) = (0, nums[n - 1] - nums[0] + 1);
let check = |diff: i32| -> bool {
let mut cnt = 0;
let mut i = 0;
while i < n - 1 {
if nums[i + 1] - nums[i] <= diff {
cnt += 1;
i += 2;
} else {
i += 1;
}
}
cnt >= p
};
while l < r {
let mid = (l + r) / 2;
if check(mid) {
r = mid;
} else {
l = mid + 1;
}
}
l
}
}
// Accepted solution for LeetCode #2616: Minimize the Maximum Difference of Pairs
function minimizeMax(nums: number[], p: number): number {
nums.sort((a, b) => a - b);
const n = nums.length;
let l = 0,
r = nums[n - 1] - nums[0] + 1;
const check = (diff: number): boolean => {
let cnt = 0;
for (let i = 0; i < n - 1; ++i) {
if (nums[i + 1] - nums[i] <= diff) {
++cnt;
++i;
}
}
return cnt >= p;
};
while (l < r) {
const mid = (l + r) >> 1;
if (check(mid)) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
Use this to step through a reusable interview workflow for this problem.
Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.
Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).
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: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.
Usually fails on: Two-element ranges never converge.
Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.
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.