LeetCode #97 — MEDIUM

Interleaving String

Move from brute-force thinking to an efficient approach using dynamic programming strategy.

The Problem

Problem Statement

Given strings s1, s2, and s3, find whether s3 is formed by an interleaving of s1 and s2.

An interleaving of two strings s and t is a configuration where s and t are divided into n and m substrings respectively, such that:

s = s₁ + s₂ + ... + s_n
t = t₁ + t₂ + ... + t_m
|n - m| <= 1
The interleaving is s₁ + t₁ + s₂ + t₂ + s₃ + t₃ + ... or t₁ + s₁ + t₂ + s₂ + t₃ + s₃ + ...

Note: a + b is the concatenation of strings a and b.

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Explanation: One way to obtain s3 is:
Split s1 into s1 = "aa" + "bc" + "c", and s2 into s2 = "dbbc" + "a".
Interleaving the two splits, we get "aa" + "dbbc" + "bc" + "a" + "c" = "aadbbcbcac".
Since s3 can be obtained by interleaving s1 and s2, we return true.

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
Explanation: Notice how it is impossible to interleave s2 with any other string to obtain s3.

Example 3:

Input: s1 = "", s2 = "", s3 = ""
Output: true

Constraints:

0 <= s1.length, s2.length <= 100
0 <= s3.length <= 200
s1, s2, and s3 consist of lowercase English letters.

Follow up: Could you solve it using only O(s2.length) additional memory space?

Step 01

Brute Force Baseline

Problem summary: Given strings s1, s2, and s3, find whether s3 is formed by an interleaving of s1 and s2. An interleaving of two strings s and t is a configuration where s and t are divided into n and m substrings respectively, such that: s = s1 + s2 + ... + sn t = t1 + t2 + ... + tm |n - m| <= 1 The interleaving is s1 + t1 + s2 + t2 + s3 + t3 + ... or t1 + s1 + t2 + s2 + t3 + s3 + ... Note: a + b is the concatenation of strings a and b.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming

Example 1

"aabcc"
"dbbca"
"aadbbcbcac"

Example 2

"aabcc"
"dbbca"
"aadbbbaccc"

Example 3

""
""
""

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        int m = s1.length(), n = s2.length();
        if (m + n != s3.length()) return false;

        boolean[] dp = new boolean[n + 1];
        dp[0] = true;
        for (int j = 1; j <= n; j++) {
            dp[j] = dp[j - 1] && s2.charAt(j - 1) == s3.charAt(j - 1);
        }

        for (int i = 1; i <= m; i++) {
            dp[0] = dp[0] && s1.charAt(i - 1) == s3.charAt(i - 1);
            for (int j = 1; j <= n; j++) {
                char want = s3.charAt(i + j - 1);
                dp[j] = (dp[j] && s1.charAt(i - 1) == want)
                    || (dp[j - 1] && s2.charAt(j - 1) == want);
            }
        }
        return dp[n];
    }
}

func isInterleave(s1 string, s2 string, s3 string) bool {
    m, n := len(s1), len(s2)
    if m+n != len(s3) {
        return false
    }

    dp := make([]bool, n+1)
    dp[0] = true
    for j := 1; j <= n; j++ {
        dp[j] = dp[j-1] && s2[j-1] == s3[j-1]
    }

    for i := 1; i <= m; i++ {
        dp[0] = dp[0] && s1[i-1] == s3[i-1]
        for j := 1; j <= n; j++ {
            want := s3[i+j-1]
            dp[j] = (dp[j] && s1[i-1] == want) || (dp[j-1] && s2[j-1] == want)
        }
    }
    return dp[n]
}

class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        m, n = len(s1), len(s2)
        if m + n != len(s3):
            return False

        dp = [False] * (n + 1)
        dp[0] = True
        for j in range(1, n + 1):
            dp[j] = dp[j - 1] and s2[j - 1] == s3[j - 1]

        for i in range(1, m + 1):
            dp[0] = dp[0] and s1[i - 1] == s3[i - 1]
            for j in range(1, n + 1):
                want = s3[i + j - 1]
                dp[j] = (dp[j] and s1[i - 1] == want) or (dp[j - 1] and s2[j - 1] == want)
        return dp[n]

impl Solution {
    pub fn is_interleave(s1: String, s2: String, s3: String) -> bool {
        let m = s1.len();
        let n = s2.len();
        if m + n != s3.len() {
            return false;
        }

        let b1 = s1.as_bytes();
        let b2 = s2.as_bytes();
        let b3 = s3.as_bytes();
        let mut dp = vec![false; n + 1];
        dp[0] = true;
        for j in 1..=n {
            dp[j] = dp[j - 1] && b2[j - 1] == b3[j - 1];
        }

        for i in 1..=m {
            dp[0] = dp[0] && b1[i - 1] == b3[i - 1];
            for j in 1..=n {
                let want = b3[i + j - 1];
                dp[j] = (dp[j] && b1[i - 1] == want) || (dp[j - 1] && b2[j - 1] == want);
            }
        }
        dp[n]
    }
}

function isInterleave(s1: string, s2: string, s3: string): boolean {
  const m = s1.length;
  const n = s2.length;
  if (m + n !== s3.length) return false;

  const dp: boolean[] = Array(n + 1).fill(false);
  dp[0] = true;
  for (let j = 1; j <= n; j++) {
    dp[j] = dp[j - 1] && s2[j - 1] === s3[j - 1];
  }

  for (let i = 1; i <= m; i++) {
    dp[0] = dp[0] && s1[i - 1] === s3[i - 1];
    for (let j = 1; j <= n; j++) {
      const want = s3[i + j - 1];
      dp[j] = (dp[j] && s1[i - 1] === want) || (dp[j - 1] && s2[j - 1] === want);
    }
  }

  return dp[n];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m × n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.