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.
You are given two 2D integer arrays nums1 and nums2.
nums1[i] = [idi, vali] indicate that the number with the id idi has a value equal to vali.nums2[i] = [idi, vali] indicate that the number with the id idi has a value equal to vali.Each array contains unique ids and is sorted in ascending order by id.
Merge the two arrays into one array that is sorted in ascending order by id, respecting the following conditions:
0.Return the resulting array. The returned array must be sorted in ascending order by id.
Example 1:
Input: nums1 = [[1,2],[2,3],[4,5]], nums2 = [[1,4],[3,2],[4,1]] Output: [[1,6],[2,3],[3,2],[4,6]] Explanation: The resulting array contains the following: - id = 1, the value of this id is 2 + 4 = 6. - id = 2, the value of this id is 3. - id = 3, the value of this id is 2. - id = 4, the value of this id is 5 + 1 = 6.
Example 2:
Input: nums1 = [[2,4],[3,6],[5,5]], nums2 = [[1,3],[4,3]] Output: [[1,3],[2,4],[3,6],[4,3],[5,5]] Explanation: There are no common ids, so we just include each id with its value in the resulting list.
Constraints:
1 <= nums1.length, nums2.length <= 200nums1[i].length == nums2[j].length == 21 <= idi, vali <= 1000Problem summary: You are given two 2D integer arrays nums1 and nums2. nums1[i] = [idi, vali] indicate that the number with the id idi has a value equal to vali. nums2[i] = [idi, vali] indicate that the number with the id idi has a value equal to vali. Each array contains unique ids and is sorted in ascending order by id. Merge the two arrays into one array that is sorted in ascending order by id, respecting the following conditions: Only ids that appear in at least one of the two arrays should be included in the resulting array. Each id should be included only once and its value should be the sum of the values of this id in the two arrays. If the id does not exist in one of the two arrays, then assume its value in that array to be 0. Return the resulting array. The returned array must be sorted in ascending order by id.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map · Two Pointers
[[1,2],[2,3],[4,5]] [[1,4],[3,2],[4,1]]
[[2,4],[3,6],[5,5]] [[1,3],[4,3]]
merge-two-sorted-lists)meeting-scheduler)merge-similar-items)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2570: Merge Two 2D Arrays by Summing Values
class Solution {
public int[][] mergeArrays(int[][] nums1, int[][] nums2) {
int[] cnt = new int[1001];
for (var x : nums1) {
cnt[x[0]] += x[1];
}
for (var x : nums2) {
cnt[x[0]] += x[1];
}
int n = 0;
for (int i = 0; i < 1001; ++i) {
if (cnt[i] > 0) {
++n;
}
}
int[][] ans = new int[n][2];
for (int i = 0, j = 0; i < 1001; ++i) {
if (cnt[i] > 0) {
ans[j++] = new int[] {i, cnt[i]};
}
}
return ans;
}
}
// Accepted solution for LeetCode #2570: Merge Two 2D Arrays by Summing Values
func mergeArrays(nums1 [][]int, nums2 [][]int) (ans [][]int) {
cnt := [1001]int{}
for _, x := range nums1 {
cnt[x[0]] += x[1]
}
for _, x := range nums2 {
cnt[x[0]] += x[1]
}
for i, x := range cnt {
if x > 0 {
ans = append(ans, []int{i, x})
}
}
return
}
# Accepted solution for LeetCode #2570: Merge Two 2D Arrays by Summing Values
class Solution:
def mergeArrays(
self, nums1: List[List[int]], nums2: List[List[int]]
) -> List[List[int]]:
cnt = Counter()
for i, v in nums1 + nums2:
cnt[i] += v
return sorted(cnt.items())
// Accepted solution for LeetCode #2570: Merge Two 2D Arrays by Summing Values
impl Solution {
pub fn merge_arrays(nums1: Vec<Vec<i32>>, nums2: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
let mut cnt = vec![0; 1001];
for x in &nums1 {
cnt[x[0] as usize] += x[1];
}
for x in &nums2 {
cnt[x[0] as usize] += x[1];
}
let mut ans = vec![];
for i in 0..cnt.len() {
if cnt[i] > 0 {
ans.push(vec![i as i32, cnt[i] as i32]);
}
}
ans
}
}
// Accepted solution for LeetCode #2570: Merge Two 2D Arrays by Summing Values
function mergeArrays(nums1: number[][], nums2: number[][]): number[][] {
const n = 1001;
const cnt = new Array(n).fill(0);
for (const [a, b] of nums1) {
cnt[a] += b;
}
for (const [a, b] of nums2) {
cnt[a] += b;
}
const ans: number[][] = [];
for (let i = 0; i < n; ++i) {
if (cnt[i] > 0) {
ans.push([i, cnt[i]]);
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
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: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.