LeetCode #1585 — HARD

Check If String Is Transformable With Substring Sort Operations

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

Given two strings s and t, transform string s into string t using the following operation any number of times:

Choose a non-empty substring in s and sort it in place so the characters are in ascending order.
- For example, applying the operation on the underlined substring in "14234" results in "12344".

Return true if it is possible to transform s into t. Otherwise, return false.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "84532", t = "34852"
Output: true
Explanation: You can transform s into t using the following sort operations:
"84532" (from index 2 to 3) -> "84352"
"84352" (from index 0 to 2) -> "34852"

Example 2:

Input: s = "34521", t = "23415"
Output: true
Explanation: You can transform s into t using the following sort operations:
"34521" -> "23451"
"23451" -> "23415"

Example 3:

Input: s = "12345", t = "12435"
Output: false

Constraints:

s.length == t.length
1 <= s.length <= 10⁵
s and t consist of only digits.

Step 01

Brute Force Baseline

Problem summary: Given two strings s and t, transform string s into string t using the following operation any number of times: Choose a non-empty substring in s and sort it in place so the characters are in ascending order. For example, applying the operation on the underlined substring in "14234" results in "12344". Return true if it is possible to transform s into t. Otherwise, return false. A substring is a contiguous sequence of characters within a string.

Baseline thinking

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

Pattern signal: Greedy

Example 1

"84532"
"34852"

Example 2

"34521"
"23415"

Example 3

"12345"
"12435"

Step 02

Core Insight

What unlocks the optimal approach

Suppose the first digit you need is 'd'. How can you determine if it's possible to get that digit there?
Consider swapping adjacent characters to maintain relative ordering.

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

Largest constraint values

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

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1585: Check If String Is Transformable With Substring Sort Operations
class Solution {
    public boolean isTransformable(String s, String t) {
        Deque<Integer>[] pos = new Deque[10];
        Arrays.setAll(pos, k -> new ArrayDeque<>());
        for (int i = 0; i < s.length(); ++i) {
            pos[s.charAt(i) - '0'].offer(i);
        }
        for (int i = 0; i < t.length(); ++i) {
            int x = t.charAt(i) - '0';
            if (pos[x].isEmpty()) {
                return false;
            }
            for (int j = 0; j < x; ++j) {
                if (!pos[j].isEmpty() && pos[j].peek() < pos[x].peek()) {
                    return false;
                }
            }
            pos[x].poll();
        }
        return true;
    }
}

// Accepted solution for LeetCode #1585: Check If String Is Transformable With Substring Sort Operations
func isTransformable(s string, t string) bool {
	pos := [10][]int{}
	for i, c := range s {
		pos[c-'0'] = append(pos[c-'0'], i)
	}
	for _, c := range t {
		x := int(c - '0')
		if len(pos[x]) == 0 {
			return false
		}
		for j := 0; j < x; j++ {
			if len(pos[j]) > 0 && pos[j][0] < pos[x][0] {
				return false
			}
		}
		pos[x] = pos[x][1:]
	}
	return true
}

# Accepted solution for LeetCode #1585: Check If String Is Transformable With Substring Sort Operations
class Solution:
    def isTransformable(self, s: str, t: str) -> bool:
        pos = defaultdict(deque)
        for i, c in enumerate(s):
            pos[int(c)].append(i)
        for c in t:
            x = int(c)
            if not pos[x] or any(pos[i] and pos[i][0] < pos[x][0] for i in range(x)):
                return False
            pos[x].popleft()
        return True

// Accepted solution for LeetCode #1585: Check If String Is Transformable With Substring Sort Operations
struct Solution;

impl Solution {
    fn is_transformable(s: String, t: String) -> bool {
        let n = s.len();
        let s: Vec<u8> = s.bytes().collect();
        let t: Vec<u8> = t.bytes().collect();
        let mut idx: Vec<Vec<usize>> = vec![vec![]; 10];
        for i in (0..n).rev() {
            idx[(s[i] - b'0') as usize].push(i);
        }
        for i in 0..n {
            let k = (t[i] - b'0') as usize;
            if idx[k].is_empty() {
                return false;
            }
            let p = idx[k].pop().unwrap();
            for j in 0..k {
                if let Some(&q) = idx[j].last() {
                    if q < p {
                        return false;
                    }
                }
            }
        }
        true
    }
}

#[test]
fn test() {
    let s = "84532".to_string();
    let t = "34852".to_string();
    let res = true;
    assert_eq!(Solution::is_transformable(s, t), res);
    let s = "34521".to_string();
    let t = "23415".to_string();
    let res = true;
    assert_eq!(Solution::is_transformable(s, t), res);
    let s = "12345".to_string();
    let t = "12435".to_string();
    let res = false;
    assert_eq!(Solution::is_transformable(s, t), res);
    let s = "1".to_string();
    let t = "2".to_string();
    let res = false;
    assert_eq!(Solution::is_transformable(s, t), res);
}

// Accepted solution for LeetCode #1585: Check If String Is Transformable With Substring Sort Operations
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1585: Check If String Is Transformable With Substring Sort Operations
// class Solution {
//     public boolean isTransformable(String s, String t) {
//         Deque<Integer>[] pos = new Deque[10];
//         Arrays.setAll(pos, k -> new ArrayDeque<>());
//         for (int i = 0; i < s.length(); ++i) {
//             pos[s.charAt(i) - '0'].offer(i);
//         }
//         for (int i = 0; i < t.length(); ++i) {
//             int x = t.charAt(i) - '0';
//             if (pos[x].isEmpty()) {
//                 return false;
//             }
//             for (int j = 0; j < x; ++j) {
//                 if (!pos[j].isEmpty() && pos[j].peek() < pos[x].peek()) {
//                     return false;
//                 }
//             }
//             pos[x].poll();
//         }
//         return true;
//     }
// }

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(n × C)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.