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97. Interleaving String

medium

Decide whether s3 is formed by interleaving s1 and s2 — preserving order within each. A natural 2D DP where dp[i][j] tracks whether the first i chars of s1 and first j chars of s2 can be interleaved into the first i+j chars of s3.

By Sam K., Founder, InterviewChamp.AI · Last verified

Problem

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, and the interleaving is s1 + t1 + s2 + t2 + ... or t1 + s1 + t2 + s2 + .... Note: a + b is the concatenation of strings a and b.

Constraints

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

Examples

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

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Output

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Hints

Progressive — try the first before opening the next.

Hint 1

If len(s1) + len(s2) != len(s3), return false immediately.

Hint 2

dp[i][j] = can the first i chars of s1 and first j chars of s2 interleave to form the first i+j chars of s3?

Hint 3

dp[i][j] is true if (s1[i-1] == s3[i+j-1] AND dp[i-1][j]) OR (s2[j-1] == s3[i+j-1] AND dp[i][j-1]).

Hint 4

Base case dp[0][0] = true.

Solution approach

Reveal approach

Bottom-up DP over a 2D boolean table dp[m+1][n+1]. dp[i][j] means the first i chars of s1 and first j chars of s2 can interleave to form the first i+j chars of s3. Initialize dp[0][0] = true; dp[i][0] = dp[i-1][0] AND s1[i-1] == s3[i-1]; dp[0][j] symmetric. For each i and j: dp[i][j] = (s1[i-1] == s3[i+j-1] AND dp[i-1][j]) OR (s2[j-1] == s3[i+j-1] AND dp[i][j-1]). Return dp[m][n]. Space optimization: only the previous row is needed — collapse to a 1D array of length n+1.

Complexity

Time
O(m * n)
Space
O(n)

Related patterns

  • dynamic-programming
  • string-dp
  • two-strings-table

Related problems

Asked at

Companies reported asking this problem (sourced from public Glassdoor, Blind, and Levels.fyi interview posts).

  • Amazon
  • Microsoft

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