LeetCode #1638 — MEDIUM

Count Substrings That Differ by One Character

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

Given two strings s and t, find the number of ways you can choose a non-empty substring of s and replace a single character by a different character such that the resulting substring is a substring of t. In other words, find the number of substrings in s that differ from some substring in t by exactly one character.

For example, the underlined substrings in "computer" and "computation" only differ by the 'e'/'a', so this is a valid way.

Return the number of substrings that satisfy the condition above.

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

Example 1:

Input: s = "aba", t = "baba"
Output: 6
Explanation: The following are the pairs of substrings from s and t that differ by exactly 1 character:
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
The underlined portions are the substrings that are chosen from s and t.

Example 2:

Input: s = "ab", t = "bb"
Output: 3
Explanation: The following are the pairs of substrings from s and t that differ by 1 character:
("ab", "bb")
("ab", "bb")
("ab", "bb")
The underlined portions are the substrings that are chosen from s and t.

Constraints:

1 <= s.length, t.length <= 100
s and t consist of lowercase English letters only.

Step 01

Brute Force Baseline

Problem summary: Given two strings s and t, find the number of ways you can choose a non-empty substring of s and replace a single character by a different character such that the resulting substring is a substring of t. In other words, find the number of substrings in s that differ from some substring in t by exactly one character. For example, the underlined substrings in "computer" and "computation" only differ by the 'e'/'a', so this is a valid way. Return the number of substrings that satisfy the condition above. 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: Hash Map · Dynamic Programming

Example 1

"aba"
"baba"

Example 2

"ab"
"bb"

Core Insight

What unlocks the optimal approach

Take every substring of s, change a character, and see how many substrings of t match that substring.
Use a Trie to store all substrings of t as a dictionary.

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

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

// Accepted solution for LeetCode #1638: Count Substrings That Differ by One Character
class Solution {
    public int countSubstrings(String s, String t) {
        int ans = 0;
        int m = s.length(), n = t.length();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (s.charAt(i) != t.charAt(j)) {
                    int l = 0, r = 0;
                    while (i - l > 0 && j - l > 0 && s.charAt(i - l - 1) == t.charAt(j - l - 1)) {
                        ++l;
                    }
                    while (i + r + 1 < m && j + r + 1 < n
                        && s.charAt(i + r + 1) == t.charAt(j + r + 1)) {
                        ++r;
                    }
                    ans += (l + 1) * (r + 1);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1638: Count Substrings That Differ by One Character
func countSubstrings(s string, t string) (ans int) {
	m, n := len(s), len(t)
	for i, a := range s {
		for j, b := range t {
			if a != b {
				l, r := 0, 0
				for i > l && j > l && s[i-l-1] == t[j-l-1] {
					l++
				}
				for i+r+1 < m && j+r+1 < n && s[i+r+1] == t[j+r+1] {
					r++
				}
				ans += (l + 1) * (r + 1)
			}
		}
	}
	return
}

# Accepted solution for LeetCode #1638: Count Substrings That Differ by One Character
class Solution:
    def countSubstrings(self, s: str, t: str) -> int:
        ans = 0
        m, n = len(s), len(t)
        for i, a in enumerate(s):
            for j, b in enumerate(t):
                if a != b:
                    l = r = 0
                    while i > l and j > l and s[i - l - 1] == t[j - l - 1]:
                        l += 1
                    while (
                        i + r + 1 < m and j + r + 1 < n and s[i + r + 1] == t[j + r + 1]
                    ):
                        r += 1
                    ans += (l + 1) * (r + 1)
        return ans

// Accepted solution for LeetCode #1638: Count Substrings That Differ by One Character
struct Solution;

impl Solution {
    fn count_substrings(s: String, t: String) -> i32 {
        let n = s.len();
        let m = t.len();
        let s: Vec<char> = s.chars().collect();
        let t: Vec<char> = t.chars().collect();
        let mut res = 0;
        for i in 0..n {
            for j in 0..m {
                let k = (n - i).min(m - j);
                let mut diff = 0;
                for d in 0..k {
                    if s[i + d] != t[j + d] {
                        diff += 1;
                    }
                    if diff == 1 {
                        res += 1;
                    }
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let s = "aba".to_string();
    let t = "baba".to_string();
    let res = 6;
    assert_eq!(Solution::count_substrings(s, t), res);
    let s = "ab".to_string();
    let t = "bb".to_string();
    let res = 3;
    assert_eq!(Solution::count_substrings(s, t), res);
    let s = "a".to_string();
    let t = "a".to_string();
    let res = 0;
    assert_eq!(Solution::count_substrings(s, t), res);
    let s = "abe".to_string();
    let t = "bbc".to_string();
    let res = 10;
    assert_eq!(Solution::count_substrings(s, t), res);
}

// Accepted solution for LeetCode #1638: Count Substrings That Differ by One Character
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1638: Count Substrings That Differ by One Character
// class Solution {
//     public int countSubstrings(String s, String t) {
//         int ans = 0;
//         int m = s.length(), n = t.length();
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 if (s.charAt(i) != t.charAt(j)) {
//                     int l = 0, r = 0;
//                     while (i - l > 0 && j - l > 0 && s.charAt(i - l - 1) == t.charAt(j - l - 1)) {
//                         ++l;
//                     }
//                     while (i + r + 1 < m && j + r + 1 < n
//                         && s.charAt(i + r + 1) == t.charAt(j + r + 1)) {
//                         ++r;
//                     }
//                     ans += (l + 1) * (r + 1);
//                 }
//             }
//         }
//         return ans;
//     }
// }

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 × m)

Space

O(n × m)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.