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.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
Given an array nums, you can perform the following operation any number of times:
nums. If multiple such pairs exist, choose the leftmost one.Return the minimum number of operations needed to make the array non-decreasing.
An array is said to be non-decreasing if each element is greater than or equal to its previous element (if it exists).
Example 1:
Input: nums = [5,2,3,1]
Output: 2
Explanation:
(3,1) has the minimum sum of 4. After replacement, nums = [5,2,4].(2,4) has the minimum sum of 6. After replacement, nums = [5,6].The array nums became non-decreasing in two operations.
Example 2:
Input: nums = [1,2,2]
Output: 0
Explanation:
The array nums is already sorted.
Constraints:
1 <= nums.length <= 50-1000 <= nums[i] <= 1000Problem summary: Given an array nums, you can perform the following operation any number of times: Select the adjacent pair with the minimum sum in nums. If multiple such pairs exist, choose the leftmost one. Replace the pair with their sum. Return the minimum number of operations needed to make the array non-decreasing. An array is said to be non-decreasing if each element is greater than or equal to its previous element (if it exists).
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map · Linked List · Segment Tree
[5,2,3,1]
[1,2,2]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3507: Minimum Pair Removal to Sort Array I
class Solution {
public int minimumPairRemoval(int[] nums) {
List<Integer> arr = new ArrayList<>();
for (int x : nums) {
arr.add(x);
}
int ans = 0;
while (!isNonDecreasing(arr)) {
int k = 0;
int s = arr.get(0) + arr.get(1);
for (int i = 1; i < arr.size() - 1; ++i) {
int t = arr.get(i) + arr.get(i + 1);
if (s > t) {
s = t;
k = i;
}
}
arr.set(k, s);
arr.remove(k + 1);
++ans;
}
return ans;
}
private boolean isNonDecreasing(List<Integer> arr) {
for (int i = 1; i < arr.size(); ++i) {
if (arr.get(i) < arr.get(i - 1)) {
return false;
}
}
return true;
}
}
// Accepted solution for LeetCode #3507: Minimum Pair Removal to Sort Array I
func minimumPairRemoval(nums []int) int {
arr := append([]int(nil), nums...)
ans := 0
isNonDecreasing := func(a []int) bool {
for i := 1; i < len(a); i++ {
if a[i] < a[i-1] {
return false
}
}
return true
}
for !isNonDecreasing(arr) {
k := 0
s := arr[0] + arr[1]
for i := 1; i < len(arr)-1; i++ {
t := arr[i] + arr[i+1]
if s > t {
s = t
k = i
}
}
arr[k] = s
copy(arr[k+1:], arr[k+2:])
arr = arr[:len(arr)-1]
ans++
}
return ans
}
# Accepted solution for LeetCode #3507: Minimum Pair Removal to Sort Array I
class Solution:
def minimumPairRemoval(self, nums: List[int]) -> int:
arr = nums[:]
ans = 0
def is_non_decreasing(a: List[int]) -> bool:
for i in range(1, len(a)):
if a[i] < a[i - 1]:
return False
return True
while not is_non_decreasing(arr):
k = 0
s = arr[0] + arr[1]
for i in range(1, len(arr) - 1):
t = arr[i] + arr[i + 1]
if s > t:
s = t
k = i
arr[k] = s
arr.pop(k + 1)
ans += 1
return ans
// Accepted solution for LeetCode #3507: Minimum Pair Removal to Sort Array I
impl Solution {
pub fn minimum_pair_removal(nums: Vec<i32>) -> i32 {
let mut arr: Vec<i32> = nums.clone();
let mut ans: i32 = 0;
fn is_non_decreasing(a: &Vec<i32>) -> bool {
for i in 1..a.len() {
if a[i] < a[i - 1] {
return false;
}
}
true
}
while !is_non_decreasing(&arr) {
let mut k: usize = 0;
let mut s: i32 = arr[0] + arr[1];
for i in 1..arr.len() - 1 {
let t: i32 = arr[i] + arr[i + 1];
if s > t {
s = t;
k = i;
}
}
arr[k] = s;
arr.remove(k + 1);
ans += 1;
}
ans
}
}
// Accepted solution for LeetCode #3507: Minimum Pair Removal to Sort Array I
function minimumPairRemoval(nums: number[]): number {
const arr = nums.slice();
let ans = 0;
const isNonDecreasing = (a: number[]): boolean => {
for (let i = 1; i < a.length; i++) {
if (a[i] < a[i - 1]) {
return false;
}
}
return true;
};
while (!isNonDecreasing(arr)) {
let k = 0;
let s = arr[0] + arr[1];
for (let i = 1; i < arr.length - 1; ++i) {
const t = arr[i] + arr[i + 1];
if (s > t) {
s = t;
k = i;
}
}
arr[k] = s;
arr.splice(k + 1, 1);
ans++;
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.
Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.
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: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.
Wrong move: Pointer updates overwrite references before they are saved.
Usually fails on: List becomes disconnected mid-operation.
Fix: Store next pointers first and use a dummy head for safer joins.