# 97. Interleaving String

## Description

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.

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?

## Solutions

Solution 1: Memoization Search

Let’s denote the length of string $s_1$ as $m$ and the length of string $s_2$ as $n$. If $m + n \neq |s_3|$, then $s_3$ is definitely not an interleaving string of $s_1$ and $s_2$, so we return false.

Next, we design a function $dfs(i, j)$, which represents whether the remaining part of $s_3$ can be interleaved from the $i$th character of $s_1$ and the $j$th character of $s_2$. The answer is $dfs(0, 0)$.

The calculation process of function $dfs(i, j)$ is as follows:

If $i \geq m$ and $j \geq n$, it means that both $s_1$ and $s_2$ have been traversed, so we return true.

If $i < m$ and $s_1[i] = s_3[i + j]$, it means that the character $s_1[i]$ is part of $s_3[i + j]$. Therefore, we recursively call $dfs(i + 1, j)$ to check whether the next character of $s_1$ can match the current character of $s_2$. If it can match, we return true.

Similarly, if $j < n$ and $s_2[j] = s_3[i + j]$, it means that the character $s_2[j]$ is part of $s_3[i + j]$. Therefore, we recursively call $dfs(i, j + 1)$ to check whether the next character of $s_2$ can match the current character of $s_1$. If it can match, we return true.

Otherwise, we return false.

To avoid repeated calculations, we can use memoization search.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of strings $s_1$ and $s_2$ respectively.

Solution 2: Dynamic Programming

We can convert the memoization search in Solution 1 into dynamic programming.

We define $f[i][j]$ to represent whether the first $i$ characters of string $s_1$ and the first $j$ characters of string $s_2$ can interleave to form the first $i + j$ characters of string $s_3$. When transitioning states, we can consider whether the current character is obtained from the last character of $s_1$ or the last character of $s_2$. Therefore, we have the state transition equation:

$f[i][j] = \begin{cases} f[i - 1][j] & \text{if } s_1[i - 1] = s_3[i + j - 1] \\ \text{or } f[i][j - 1] & \text{if } s_2[j - 1] = s_3[i + j - 1] \\ \text{false} & \text{otherwise} \end{cases}$

where $f[0][0] = \text{true}$ indicates that an empty string is an interleaving string of two empty strings.

The answer is $f[m][n]$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of strings $s_1$ and $s_2$ respectively.

We notice that the state $f[i][j]$ is only related to the states $f[i - 1][j]$, $f[i][j - 1]$, and $f[i - 1][j - 1]$. Therefore, we can use a rolling array to optimize the space complexity, reducing the original space complexity from $O(m \times n)$ to $O(n)$.

• 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[] f = new boolean[n + 1];
f[0] = true;
for (int i = 0; i <= m; ++i) {
for (int j = 0; j <= n; ++j) {
int k = i + j - 1;
if (i > 0) {
f[j] &= s1.charAt(i - 1) == s3.charAt(k);
}
if (j > 0) {
f[j] |= (f[j - 1] & s2.charAt(j - 1) == s3.charAt(k));
}
}
}
return f[n];
}
}

//////

public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if (s1.length() == 0 || s1 == null) {
return s2.equals(s3);
}

if (s2.length() == 0 || s2 == null) {
return s1.equals(s3);
}

// @note: missed this simple check
if (s1.length() + s2.length() != s3.length()) {
return false;
}

// 1-to-i of s1, and 1-to-j of s2, can for 1-to-(i+j) of s3
boolean[][] dp = new boolean[s1.length() + 1][s2.length() + 1]; // +1 for empty string

dp[0][0] = true;

// pre-process for empty string case
for (int i = 1; i < s1.length() + 1; i++) {
dp[i][0] = (s1.charAt(i - 1) == s3.charAt(i - 1))
&& dp[i - 1][0];
}
for (int i = 1; i < s2.length() + 1; i++) {
dp[0][i] = (s2.charAt(i - 1) == s3.charAt(i - 1))
&& dp[0][i - 1];
}

for (int i = 1; i < s1.length() + 1; i++) {
for (int j = 1; j < s2.length() + 1; j++) {

boolean withS1 = s1.charAt(i - 1) == s3.charAt(i - 1 + j) && dp[i - 1][j];
boolean withS2 = s2.charAt(j - 1) == s3.charAt(i - 1 + j) && dp[i][j - 1];

dp[i][j] = withS1 || withS2;
}
}

return dp[s1.length()][s2.length()];
}
}

• class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
int m = s1.size(), n = s2.size();
if (m + n != s3.size()) {
return false;
}
bool f[n + 1];
memset(f, false, sizeof(f));
f[0] = true;
for (int i = 0; i <= m; ++i) {
for (int j = 0; j <= n; ++j) {
int k = i + j - 1;
if (i) {
f[j] &= s1[i - 1] == s3[k];
}
if (j) {
f[j] |= (s2[j - 1] == s3[k] && f[j - 1]);
}
}
}
return f[n];
}
};

• # 2-d
class Solution:
def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
if len(s1) + len(s2) != len(s3):
return False

dp = [[False] * (len(s2) + 1) for _ in range(len(s1) + 1)]
dp[0][0] = True

for i in range(1, len(s1) + 1):
dp[i][0] = s1[i - 1] == s3[i - 1] and dp[i - 1][0]
for i in range(1, len(s2) + 1):
dp[0][i] = s2[i - 1] == s3[i - 1] and dp[0][i - 1]

# fill in dp array
for i in range(1, len(s1) + 1):
for j in range(1, len(s2) + 1):
with_s1 = s1[i - 1] == s3[i + j - 1] and dp[i - 1][j]
with_s2 = s2[j - 1] == s3[i + j - 1] and dp[i][j - 1]
dp[i][j] = with_s1 or with_s2

return dp[-1][-1]

############

# 1-d
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
f = [True] + [False] * n
for i in range(m + 1):
for j in range(n + 1):
k = i + j - 1
if i:
f[j] &= s1[i - 1] == s3[k]
if j:
f[j] |= f[j - 1] and s2[j - 1] == s3[k]
return f[n]

###########

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

@cache
def dfs(i, j):
if i == m and j == n:
return True

return (
( i < m and s1[i] == s3[i + j] and dfs(i + 1, j) )
or
( j < n and s2[j] == s3[i + j] and dfs(i, j + 1) )
)

return dfs(0, 0)

• func isInterleave(s1 string, s2 string, s3 string) bool {
m, n := len(s1), len(s2)
if m+n != len(s3) {
return false
}
f := make([]bool, n+1)
f[0] = true
for i := 0; i <= m; i++ {
for j := 0; j <= n; j++ {
k := i + j - 1
if i > 0 {
f[j] = (f[j] && s1[i-1] == s3[k])
}
if j > 0 {
f[j] = (f[j] || (s2[j-1] == s3[k] && f[j-1]))
}
}
}
return f[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 f: boolean[] = new Array(n + 1).fill(false);
f[0] = true;
for (let i = 0; i <= m; ++i) {
for (let j = 0; j <= n; ++j) {
const k = i + j - 1;
if (i) {
f[j] = f[j] && s1[i - 1] === s3[k];
}
if (j) {
f[j] = f[j] || (f[j - 1] && s2[j - 1] === s3[k]);
}
}
}
return f[n];
}

• public class Solution {
public bool IsInterleave(string s1, string s2, string s3) {
int m = s1.Length, n = s2.Length;
if (m + n != s3.Length) {
return false;
}
bool[] f = new bool[n + 1];
f[0] = true;
for (int i = 0; i <= m; ++i) {
for (int j = 0; j <= n; ++j) {
int k = i + j - 1;
if (i > 0) {
f[j] &= s1[i - 1] == s3[k];
}
if (j > 0) {
f[j] |= (f[j - 1] & s2[j - 1] == s3[k]);
}
}
}
return f[n];
}
}

• impl Solution {
pub fn is_interleave(s1: String, s2: String, s3: String) -> bool {
let n = s1.len();
let m = s2.len();

if s1.len() + s2.len() != s3.len() {
return false;
}

let mut record = vec![vec![-1; m + 1]; n + 1];

Self::dfs(
&mut record,
n,
m,
0,
0,
&s1.chars().collect(),
&s2.chars().collect(),
&s3.chars().collect()
)
}

fn dfs(
record: &mut Vec<Vec<i32>>,
n: usize,
m: usize,
i: usize,
j: usize,
s1: &Vec<char>,
s2: &Vec<char>,
s3: &Vec<char>
) -> bool {
if i >= n && j >= m {
return true;
}

if record[i][j] != -1 {
return record[i][j] == 1;
}

// Set the initial value
record[i][j] = 0;
let k = i + j;

// Let's try s1 first
if i < n && s1[i] == s3[k] && Self::dfs(record, n, m, i + 1, j, s1, s2, s3) {
record[i][j] = 1;
}

// If the first approach does not succeed, let's then try s2
if
record[i][j] == 0 &&
j < m &&
s2[j] == s3[k] &&
Self::dfs(record, n, m, i, j + 1, s1, s2, s3)
{
record[i][j] = 1;
}

record[i][j] == 1
}
}