Approach Summary
Optimal solution for Merge sorted arrays with O(m+n) time complexity and O(1) space complexity.
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(2^n) | O(n) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(m+n) | O(1) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Approach 1: Baseline Brute Force
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int p1=m-1;
int p2=n-1;
int i=m+n-1;
while(p2>=0){
if(p1 >= 0 && nums1[p1] > nums2[p2]){
nums1[i] = nums1[p1];
i--;
p1--;
}
else{
nums1[i] = nums2[p2];
i--;
p2--;
}
}
}
}class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int p1=m-1;
int p2=n-1;
int i=m+n-1;
while(p2>=0){
if(p1>=0 && nums1[p1] > nums2[p2]){
nums1[i] = nums1[p1];
i--;
p1--;
}
else{
nums1[i] = nums2[p2];
i--;
p2--;
}
}
}
};class Solution:
def merge_sorted_array(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Merge sorted arrays
*/
function mergeSortedArray(...args) {
// JavaScript implementation matching the optimal algorithm
} Approach 2: Optimal Interview Solution
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int p1=m-1;
int p2=n-1;
int i=m+n-1;
while(p2>=0){
if(p1 >= 0 && nums1[p1] > nums2[p2]){
nums1[i] = nums1[p1];
i--;
p1--;
}
else{
nums1[i] = nums2[p2];
i--;
p2--;
}
}
}
}class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int p1=m-1;
int p2=n-1;
int i=m+n-1;
while(p2>=0){
if(p1>=0 && nums1[p1] > nums2[p2]){
nums1[i] = nums1[p1];
i--;
p1--;
}
else{
nums1[i] = nums2[p2];
i--;
p2--;
}
}
}
};class Solution:
def merge_sorted_array(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Merge sorted arrays
*/
function mergeSortedArray(...args) {
// JavaScript implementation matching the optimal algorithm
} Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Master the Binary Search Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Binary Search pattern.
How to Recognize This Pattern
- Common Array interview pattern
- Optimal time: O(m+n)
Complexity Analysis
Time Complexity
O(m+n)
Space Complexity
O(1)