Approach Summary
dp[i][j] = true if s1[0..i) and s2[0..j) interleave to form s3[0..i+j). Transition: use s1[i-1] or s2[j-1] if it matches s3[i+j-1].
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Interleaving String optimally, we apply the dynamic programming algorithmic strategy. dp[i][j] = true if s1[0..i) and s2[0..j) interleave to form s3[0..i+j). Transition: use s1[i-1] or s2[j-1] if it matches s3[i+j-1]. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(m × n) time, utilizing O(m × n) auxiliary memory.
Time Complexity: O(m × n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(m × n) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
class Solution {
// Time: O(m × n), Space: O(m × n)
public int interleavingString(int n) {
if (n <= 1) return n;
int prev = 0, curr = 1;
for (int i = 2; i <= n; i++) {
int temp = curr;
curr = prev + curr;
prev = temp;
}
return curr;
}
}#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Time: O(m × n), Space: O(m × n)
int interleavingString(int n) {
if (n <= 1) return n;
int prev = 0, curr = 1;
for (int i = 2; i <= n; ++i) {
int temp = curr;
curr = prev + curr;
prev = temp;
}
return curr;
}
};class Solution:
def interleaving_string(self, n, *args) -> int:
# Optimal DP approach
# Time: O(m × n), Space: O(m × n)
if isinstance(n, list):
m = len(n)
dp = [0] * (m + 1)
for i in range(1, m + 1):
dp[i] = max(dp[i - 1], dp[i - 1] + n[i - 1])
return dp[m]
dp = [0] * (n + 1)
if n >= 1: dp[1] = 1
for i in range(2, n + 1):
dp[i] = dp[i - 1] + dp[i - 2]
return dp[n]var interleavingString = function(n, ...args) {
if (Array.isArray(n)) {
const m = n.length;
const dp = new Array(m + 1).fill(0);
for (let i = 1; i <= m; i++) {
dp[i] = Math.max(dp[i - 1], dp[i - 1] + n[i - 1]);
}
return dp[m];
}
if (n <= 1) return n;
let prev = 0, curr = 1;
for (let i = 2; i <= n; i++) {
const temp = curr;
curr = prev + curr;
prev = temp;
}
return curr;
};Edge Cases to Watch
- Key recognition trigger: "Does s3 = interleave of s1 and s2".
- Constraint condition: 2D DP on lengths — each cell extends one string or the other.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(m × n) time budget without stack overflow or time limit exceeded.
Master the Dynamic Programming Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Dynamic Programming pattern.
How to Recognize This Pattern
- "Does s3 = interleave of s1 and s2"
- 2D DP on lengths — each cell extends one string or the other
Complexity Analysis
Time Complexity
O(m × n)
Space Complexity
O(m × n)