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0088. Merge Sorted Array

Problem
Leetcode 88. Merge Sorted Array
Difficulty 🟢 Easy
Tags Array, Two Pointers, Sorting

You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.

Merge nums1 and nums2 into a single array sorted in non-decreasing order.

The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

Examples

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]

Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]

Constraints

  • nums1.length == m + n
  • nums2.length == n
  • 0 <= m, n <= 200
  • 1 <= m + n <= 200
  • -10^9 <= nums1[i], nums2[j] <= 10^9

Approach: Two Pointers from the End

Idea

Walk from the back: place the larger of nums1[i] and nums2[j] at nums1[k] where k = m + n - 1. Decrement accordingly. After exhausting nums2 we are done; remaining nums1 already in place.

Algorithm

  1. i = m - 1, j = n - 1, k = m + n - 1.
  2. While j >= 0:
  3. If i >= 0 && nums1[i] > nums2[j]: nums1[k] = nums1[i]; i--.
  4. Else: nums1[k] = nums2[j]; j--.
  5. k--.

Complexity

  • Time: O(m + n)
  • Space: O(1)

Implementation

Go

func merge(nums1 []int, m int, nums2 []int, n int) {
    i, j, k := m-1, n-1, m+n-1
    for j >= 0 {
        if i >= 0 && nums1[i] > nums2[j] {
            nums1[k] = nums1[i]
            i--
        } else {
            nums1[k] = nums2[j]
            j--
        }
        k--
    }
}

Java

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int i = m - 1, j = n - 1, k = m + n - 1;
        while (j >= 0) {
            if (i >= 0 && nums1[i] > nums2[j]) {
                nums1[k--] = nums1[i--];
            } else {
                nums1[k--] = nums2[j--];
            }
        }
    }
}

Python

class Solution:
    def merge(self, nums1, m, nums2, n):
        i, j, k = m - 1, n - 1, m + n - 1
        while j >= 0:
            if i >= 0 and nums1[i] > nums2[j]:
                nums1[k] = nums1[i]; i -= 1
            else:
                nums1[k] = nums2[j]; j -= 1
            k -= 1

Edge Cases

  • m = 0: nums1 is all zeros → just copy nums2.
  • n = 0: do nothing.
  • All nums2 < all nums1: shifts everything right.
  • All nums2 > all nums1: append.

Visual Animation

animation.html