LeetCode #1208 — MEDIUM

Get Equal Substrings Within Budget

Move from brute-force thinking to an efficient approach using binary search strategy.

The Problem

Problem Statement

You are given two strings s and t of the same length and an integer maxCost.

You want to change s to t. Changing the i^th character of s to i^th character of t costs |s[i] - t[i]| (i.e., the absolute difference between the ASCII values of the characters).

Return the maximum length of a substring of s that can be changed to be the same as the corresponding substring of t with a cost less than or equal to maxCost. If there is no substring from s that can be changed to its corresponding substring from t, return 0.

Example 1:

Input: s = "abcd", t = "bcdf", maxCost = 3
Output: 3
Explanation: "abc" of s can change to "bcd".
That costs 3, so the maximum length is 3.

Example 2:

Input: s = "abcd", t = "cdef", maxCost = 3
Output: 1
Explanation: Each character in s costs 2 to change to character in t,  so the maximum length is 1.

Example 3:

Input: s = "abcd", t = "acde", maxCost = 0
Output: 1
Explanation: You cannot make any change, so the maximum length is 1.

Constraints:

1 <= s.length <= 10⁵
t.length == s.length
0 <= maxCost <= 10⁶
s and t consist of only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given two strings s and t of the same length and an integer maxCost. You want to change s to t. Changing the ith character of s to ith character of t costs |s[i] - t[i]| (i.e., the absolute difference between the ASCII values of the characters). Return the maximum length of a substring of s that can be changed to be the same as the corresponding substring of t with a cost less than or equal to maxCost. If there is no substring from s that can be changed to its corresponding substring from t, return 0.

Baseline thinking

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

Pattern signal: Binary Search · Sliding Window

Example 1

"abcd"
"bcdf"
3

Example 2

"abcd"
"cdef"
3

Example 3

"abcd"
"acde"
0

Core Insight

What unlocks the optimal approach

Calculate the differences between s[i] and t[i].
Use a sliding window to track the longest valid substring.

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 #1208: Get Equal Substrings Within Budget
class Solution {
    private int maxCost;
    private int[] f;
    private int n;

    public int equalSubstring(String s, String t, int maxCost) {
        n = s.length();
        f = new int[n + 1];
        this.maxCost = maxCost;
        for (int i = 0; i < n; ++i) {
            int x = Math.abs(s.charAt(i) - t.charAt(i));
            f[i + 1] = f[i] + x;
        }
        int l = 0, r = n;
        while (l < r) {
            int mid = (l + r + 1) >>> 1;
            if (check(mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }

    private boolean check(int x) {
        for (int i = 0; i + x - 1 < n; ++i) {
            int j = i + x - 1;
            if (f[j + 1] - f[i] <= maxCost) {
                return true;
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #1208: Get Equal Substrings Within Budget
func equalSubstring(s string, t string, maxCost int) int {
	n := len(s)
	f := make([]int, n+1)
	for i, a := range s {
		f[i+1] = f[i] + abs(int(a)-int(t[i]))
	}
	check := func(x int) bool {
		for i := 0; i+x-1 < n; i++ {
			if f[i+x]-f[i] <= maxCost {
				return true
			}
		}
		return false
	}
	l, r := 0, n
	for l < r {
		mid := (l + r + 1) >> 1
		if check(mid) {
			l = mid
		} else {
			r = mid - 1
		}
	}
	return l
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1208: Get Equal Substrings Within Budget
class Solution:
    def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
        def check(x):
            for i in range(n):
                j = i + mid - 1
                if j < n and f[j + 1] - f[i] <= maxCost:
                    return True
            return False

        n = len(s)
        f = list(accumulate((abs(ord(a) - ord(b)) for a, b in zip(s, t)), initial=0))
        l, r = 0, n
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return l

// Accepted solution for LeetCode #1208: Get Equal Substrings Within Budget
struct Solution;

impl Solution {
    fn equal_substring(s: String, t: String, max_cost: i32) -> i32 {
        let s: Vec<char> = s.chars().collect();
        let t: Vec<char> = t.chars().collect();
        let n = s.len();
        let mut cost: Vec<i32> = vec![0; n];
        for i in 0..n {
            cost[i] = (s[i] as i32 - t[i] as i32).abs();
        }
        let mut start = 0;
        let mut end = 0;
        let mut res = 0;
        let mut sum = 0;
        while end < n {
            if sum <= max_cost {
                sum += cost[end];
                end += 1;
            } else {
                sum -= cost[start];
                start += 1;
            }
            if sum <= max_cost {
                res = res.max(end - start);
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let s = "abcd".to_string();
    let t = "cdef".to_string();
    let max_cost = 3;
    let res = 1;
    assert_eq!(Solution::equal_substring(s, t, max_cost), res);
    let s = "abcd".to_string();
    let t = "acde".to_string();
    let max_cost = 0;
    let res = 1;
    assert_eq!(Solution::equal_substring(s, t, max_cost), res);
}

// Accepted solution for LeetCode #1208: Get Equal Substrings Within Budget
function equalSubstring(s: string, t: string, maxCost: number): number {
    const n = s.length;
    const f = Array(n + 1).fill(0);

    for (let i = 0; i < n; i++) {
        f[i + 1] = f[i] + Math.abs(s.charCodeAt(i) - t.charCodeAt(i));
    }

    const check = (x: number): boolean => {
        for (let i = 0; i + x - 1 < n; i++) {
            if (f[i + x] - f[i] <= maxCost) {
                return true;
            }
        }
        return false;
    };

    let l = 0,
        r = n;
    while (l < r) {
        const mid = (l + r + 1) >> 1;
        if (check(mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }

    return l;
}

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 × log n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.