LeetCode #833 — MEDIUM

Find And Replace in String

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

The Problem

Problem Statement

You are given a 0-indexed string s that you must perform k replacement operations on. The replacement operations are given as three 0-indexed parallel arrays, indices, sources, and targets, all of length k.

To complete the i^th replacement operation:

Check if the substring sources[i] occurs at index indices[i] in the original string s.
If it does not occur, do nothing.
Otherwise if it does occur, replace that substring with targets[i].

For example, if s = "abcd", indices[i] = 0, sources[i] = "ab", and targets[i] = "eee", then the result of this replacement will be "eeecd".

All replacement operations must occur simultaneously, meaning the replacement operations should not affect the indexing of each other. The testcases will be generated such that the replacements will not overlap.

For example, a testcase with s = "abc", indices = [0, 1], and sources = ["ab","bc"] will not be generated because the "ab" and "bc" replacements overlap.

Return the resulting string after performing all replacement operations on s.

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

Example 1:

Input: s = "abcd", indices = [0, 2], sources = ["a", "cd"], targets = ["eee", "ffff"]
Output: "eeebffff"
Explanation:
"a" occurs at index 0 in s, so we replace it with "eee".
"cd" occurs at index 2 in s, so we replace it with "ffff".

Example 2:

Input: s = "abcd", indices = [0, 2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation:
"ab" occurs at index 0 in s, so we replace it with "eee".
"ec" does not occur at index 2 in s, so we do nothing.

Constraints:

1 <= s.length <= 1000
k == indices.length == sources.length == targets.length
1 <= k <= 100
0 <= indexes[i] < s.length
1 <= sources[i].length, targets[i].length <= 50
s consists of only lowercase English letters.
sources[i] and targets[i] consist of only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed string s that you must perform k replacement operations on. The replacement operations are given as three 0-indexed parallel arrays, indices, sources, and targets, all of length k. To complete the ith replacement operation: Check if the substring sources[i] occurs at index indices[i] in the original string s. If it does not occur, do nothing. Otherwise if it does occur, replace that substring with targets[i]. For example, if s = "abcd", indices[i] = 0, sources[i] = "ab", and targets[i] = "eee", then the result of this replacement will be "eeecd". All replacement operations must occur simultaneously, meaning the replacement operations should not affect the indexing of each other. The testcases will be generated such that the replacements will not overlap. For example, a testcase with s = "abc", indices = [0, 1], and sources = ["ab","bc"] will not be generated

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

"abcd"
[0, 2]
["a", "cd"]
["eee", "ffff"]

Example 2

"abcd"
[0, 2]
["ab","ec"]
["eee","ffff"]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #833: Find And Replace in String
class Solution {
    public String findReplaceString(String s, int[] indices, String[] sources, String[] targets) {
        int n = s.length();
        var d = new int[n];
        Arrays.fill(d, -1);
        for (int k = 0; k < indices.length; ++k) {
            int i = indices[k];
            if (s.startsWith(sources[k], i)) {
                d[i] = k;
            }
        }
        var ans = new StringBuilder();
        for (int i = 0; i < n;) {
            if (d[i] >= 0) {
                ans.append(targets[d[i]]);
                i += sources[d[i]].length();
            } else {
                ans.append(s.charAt(i++));
            }
        }
        return ans.toString();
    }
}

// Accepted solution for LeetCode #833: Find And Replace in String
func findReplaceString(s string, indices []int, sources []string, targets []string) string {
	n := len(s)
	d := make([]int, n)
	for k, i := range indices {
		if strings.HasPrefix(s[i:], sources[k]) {
			d[i] = k + 1
		}
	}
	ans := &strings.Builder{}
	for i := 0; i < n; {
		if d[i] > 0 {
			ans.WriteString(targets[d[i]-1])
			i += len(sources[d[i]-1])
		} else {
			ans.WriteByte(s[i])
			i++
		}
	}
	return ans.String()
}

# Accepted solution for LeetCode #833: Find And Replace in String
class Solution:
    def findReplaceString(
        self, s: str, indices: List[int], sources: List[str], targets: List[str]
    ) -> str:
        n = len(s)
        d = [-1] * n
        for k, (i, src) in enumerate(zip(indices, sources)):
            if s.startswith(src, i):
                d[i] = k
        ans = []
        i = 0
        while i < n:
            if ~d[i]:
                ans.append(targets[d[i]])
                i += len(sources[d[i]])
            else:
                ans.append(s[i])
                i += 1
        return "".join(ans)

// Accepted solution for LeetCode #833: Find And Replace in String
struct Solution;

type Pair = (usize, usize);

impl Solution {
    fn find_replace_string(
        mut s: String,
        indexes: Vec<i32>,
        sources: Vec<String>,
        targets: Vec<String>,
    ) -> String {
        let n = indexes.len();
        let mut v: Vec<Pair> = vec![];
        for i in 0..n {
            v.push((indexes[i] as usize, i));
        }
        v.sort_unstable();
        for (i, j) in v.into_iter().rev() {
            let source = &sources[j];
            let m = source.len();
            let target = &targets[j];
            if i + m <= s.len() && &s[i..i + m] == source {
                s.replace_range(i..i + m, target);
            }
        }
        s
    }
}

#[test]
fn test() {
    let s = "abcd".to_string();
    let indexes = vec![0, 2];
    let sources = vec_string!["a", "cd"];
    let targets = vec_string!["eee", "ffff"];
    let res = "eeebffff".to_string();
    assert_eq!(
        Solution::find_replace_string(s, indexes, sources, targets),
        res
    );
    let s = "abcd".to_string();
    let indexes = vec![0, 2];
    let sources = vec_string!["ab", "ec"];
    let targets = vec_string!["eee", "ffff"];
    let res = "eeecd".to_string();
    assert_eq!(
        Solution::find_replace_string(s, indexes, sources, targets),
        res
    );
}

// Accepted solution for LeetCode #833: Find And Replace in String
function findReplaceString(
    s: string,
    indices: number[],
    sources: string[],
    targets: string[],
): string {
    const n = s.length;
    const d: number[] = Array(n).fill(-1);
    for (let k = 0; k < indices.length; ++k) {
        const [i, src] = [indices[k], sources[k]];
        if (s.startsWith(src, i)) {
            d[i] = k;
        }
    }
    const ans: string[] = [];
    for (let i = 0; i < n; ) {
        if (d[i] >= 0) {
            ans.push(targets[d[i]]);
            i += sources[d[i]].length;
        } else {
            ans.push(s[i++]);
        }
    }
    return ans.join('');
}

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(L)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.