Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
We can scramble a string s to get a string t using the following algorithm:
s, divide it to x and y where s = x + y.s may become s = x + y or s = y + x.x and y.Given two strings s1 and s2 of the same length, return true if s2 is a scrambled string of s1, otherwise, return false.
Example 1:
Input: s1 = "great", s2 = "rgeat" Output: true Explanation: One possible scenario applied on s1 is: "great" --> "gr/eat" // divide at random index. "gr/eat" --> "gr/eat" // random decision is not to swap the two substrings and keep them in order. "gr/eat" --> "g/r / e/at" // apply the same algorithm recursively on both substrings. divide at random index each of them. "g/r / e/at" --> "r/g / e/at" // random decision was to swap the first substring and to keep the second substring in the same order. "r/g / e/at" --> "r/g / e/ a/t" // again apply the algorithm recursively, divide "at" to "a/t". "r/g / e/ a/t" --> "r/g / e/ a/t" // random decision is to keep both substrings in the same order. The algorithm stops now, and the result string is "rgeat" which is s2. As one possible scenario led s1 to be scrambled to s2, we return true.
Example 2:
Input: s1 = "abcde", s2 = "caebd" Output: false
Example 3:
Input: s1 = "a", s2 = "a" Output: true
Constraints:
s1.length == s2.length1 <= s1.length <= 30s1 and s2 consist of lowercase English letters.Problem summary: We can scramble a string s to get a string t using the following algorithm: If the length of the string is 1, stop. If the length of the string is > 1, do the following: Split the string into two non-empty substrings at a random index, i.e., if the string is s, divide it to x and y where s = x + y. Randomly decide to swap the two substrings or to keep them in the same order. i.e., after this step, s may become s = x + y or s = y + x. Apply step 1 recursively on each of the two substrings x and y. Given two strings s1 and s2 of the same length, return true if s2 is a scrambled string of s1, otherwise, return false.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
"great" "rgeat"
"abcde" "caebd"
"a" "a"
import java.util.*;
class Solution {
private final Map<String, Boolean> memo = new HashMap<>();
public boolean isScramble(String s1, String s2) {
return dfs(s1, s2);
}
private boolean dfs(String a, String b) {
String key = a + "|" + b;
if (memo.containsKey(key)) return memo.get(key);
if (a.equals(b)) return true;
if (!sameCharMultiset(a, b)) {
memo.put(key, false);
return false;
}
int n = a.length();
for (int i = 1; i < n; i++) {
if (dfs(a.substring(0, i), b.substring(0, i))
&& dfs(a.substring(i), b.substring(i))) {
memo.put(key, true);
return true;
}
if (dfs(a.substring(0, i), b.substring(n - i))
&& dfs(a.substring(i), b.substring(0, n - i))) {
memo.put(key, true);
return true;
}
}
memo.put(key, false);
return false;
}
private boolean sameCharMultiset(String a, String b) {
if (a.length() != b.length()) return false;
int[] count = new int[26];
for (int i = 0; i < a.length(); i++) {
count[a.charAt(i) - 'a']++;
count[b.charAt(i) - 'a']--;
}
for (int x : count) if (x != 0) return false;
return true;
}
}
func isScramble(s1 string, s2 string) bool {
memo := map[string]bool{}
seen := map[string]bool{}
var sameChars func(a, b string) bool
sameChars = func(a, b string) bool {
if len(a) != len(b) {
return false
}
cnt := [26]int{}
for i := 0; i < len(a); i++ {
cnt[a[i]-'a']++
cnt[b[i]-'a']--
}
for _, v := range cnt {
if v != 0 {
return false
}
}
return true
}
var dfs func(a, b string) bool
dfs = func(a, b string) bool {
key := a + "|" + b
if seen[key] {
return memo[key]
}
seen[key] = true
if a == b {
memo[key] = true
return true
}
if !sameChars(a, b) {
memo[key] = false
return false
}
n := len(a)
for i := 1; i < n; i++ {
if dfs(a[:i], b[:i]) && dfs(a[i:], b[i:]) {
memo[key] = true
return true
}
if dfs(a[:i], b[n-i:]) && dfs(a[i:], b[:n-i]) {
memo[key] = true
return true
}
}
memo[key] = false
return false
}
return dfs(s1, s2)
}
class Solution:
def isScramble(self, s1: str, s2: str) -> bool:
memo = {}
def same_chars(a: str, b: str) -> bool:
if len(a) != len(b):
return False
cnt = [0] * 26
for x, y in zip(a, b):
cnt[ord(x) - 97] += 1
cnt[ord(y) - 97] -= 1
return all(v == 0 for v in cnt)
def dfs(a: str, b: str) -> bool:
key = (a, b)
if key in memo:
return memo[key]
if a == b:
memo[key] = True
return True
if not same_chars(a, b):
memo[key] = False
return False
n = len(a)
for i in range(1, n):
if dfs(a[:i], b[:i]) and dfs(a[i:], b[i:]):
memo[key] = True
return True
if dfs(a[:i], b[n - i:]) and dfs(a[i:], b[:n - i]):
memo[key] = True
return True
memo[key] = False
return False
return dfs(s1, s2)
use std::collections::HashMap;
impl Solution {
pub fn is_scramble(s1: String, s2: String) -> bool {
fn same_chars(a: &str, b: &str) -> bool {
if a.len() != b.len() {
return false;
}
let mut cnt = [0i32; 26];
let ba = a.as_bytes();
let bb = b.as_bytes();
for i in 0..ba.len() {
cnt[(ba[i] - b'a') as usize] += 1;
cnt[(bb[i] - b'a') as usize] -= 1;
}
cnt.iter().all(|&x| x == 0)
}
fn dfs(a: &str, b: &str, memo: &mut HashMap<String, bool>) -> bool {
let key = format!("{}|{}", a, b);
if let Some(&v) = memo.get(&key) {
return v;
}
if a == b {
memo.insert(key, true);
return true;
}
if !same_chars(a, b) {
memo.insert(key, false);
return false;
}
let n = a.len();
for i in 1..n {
if (dfs(&a[..i], &b[..i], memo) && dfs(&a[i..], &b[i..], memo))
|| (dfs(&a[..i], &b[n - i..], memo) && dfs(&a[i..], &b[..n - i], memo))
{
memo.insert(key, true);
return true;
}
}
memo.insert(key, false);
false
}
let mut memo = HashMap::new();
dfs(&s1, &s2, &mut memo)
}
}
function isScramble(s1: string, s2: string): boolean {
const memo = new Map<string, boolean>();
const sameChars = (a: string, b: string): boolean => {
if (a.length !== b.length) return false;
const cnt = Array(26).fill(0);
for (let i = 0; i < a.length; i++) {
cnt[a.charCodeAt(i) - 97]++;
cnt[b.charCodeAt(i) - 97]--;
}
return cnt.every((x) => x === 0);
};
const dfs = (a: string, b: string): boolean => {
const key = a + '|' + b;
if (memo.has(key)) return memo.get(key)!;
if (a === b) {
memo.set(key, true);
return true;
}
if (!sameChars(a, b)) {
memo.set(key, false);
return false;
}
const n = a.length;
for (let i = 1; i < n; i++) {
if ((dfs(a.slice(0, i), b.slice(0, i)) && dfs(a.slice(i), b.slice(i))) ||
(dfs(a.slice(0, i), b.slice(n - i)) && dfs(a.slice(i), b.slice(0, n - i)))) {
memo.set(key, true);
return true;
}
}
memo.set(key, false);
return false;
};
return dfs(s1, s2);
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.