LeetCode #2381 — MEDIUM

Shifting Letters II

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

The Problem

Problem Statement

You are given a string s of lowercase English letters and a 2D integer array shifts where shifts[i] = [start_i, end_i, direction_i]. For every i, shift the characters in s from the index start_i to the index end_i (inclusive) forward if direction_i = 1, or shift the characters backward if direction_i = 0.

Shifting a character forward means replacing it with the next letter in the alphabet (wrapping around so that 'z' becomes 'a'). Similarly, shifting a character backward means replacing it with the previous letter in the alphabet (wrapping around so that 'a' becomes 'z').

Return the final string after all such shifts to s are applied.

Example 1:

Input: s = "abc", shifts = [[0,1,0],[1,2,1],[0,2,1]]
Output: "ace"
Explanation: Firstly, shift the characters from index 0 to index 1 backward. Now s = "zac".
Secondly, shift the characters from index 1 to index 2 forward. Now s = "zbd".
Finally, shift the characters from index 0 to index 2 forward. Now s = "ace".

Example 2:

Input: s = "dztz", shifts = [[0,0,0],[1,1,1]]
Output: "catz"
Explanation: Firstly, shift the characters from index 0 to index 0 backward. Now s = "cztz".
Finally, shift the characters from index 1 to index 1 forward. Now s = "catz".

Constraints:

1 <= s.length, shifts.length <= 5 * 10⁴
shifts[i].length == 3
0 <= start_i <= end_i < s.length
0 <= direction_i <= 1
s consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s of lowercase English letters and a 2D integer array shifts where shifts[i] = [starti, endi, directioni]. For every i, shift the characters in s from the index starti to the index endi (inclusive) forward if directioni = 1, or shift the characters backward if directioni = 0. Shifting a character forward means replacing it with the next letter in the alphabet (wrapping around so that 'z' becomes 'a'). Similarly, shifting a character backward means replacing it with the previous letter in the alphabet (wrapping around so that 'a' becomes 'z'). Return the final string after all such shifts to s are applied.

Baseline thinking

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

Pattern signal: Array

Example 1

"abc"
[[0,1,0],[1,2,1],[0,2,1]]

Example 2

"dztz"
[[0,0,0],[1,1,1]]

Core Insight

What unlocks the optimal approach

Instead of shifting every character in each shift, could you keep track of which characters are shifted and by how much across all shifts?
Try marking the start and ends of each shift, then perform a prefix sum of the shifts.

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 #2381: Shifting Letters II
class Solution {
    public String shiftingLetters(String s, int[][] shifts) {
        int n = s.length();
        int[] d = new int[n + 1];
        for (int[] e : shifts) {
            if (e[2] == 0) {
                e[2]--;
            }
            d[e[0]] += e[2];
            d[e[1] + 1] -= e[2];
        }
        for (int i = 1; i <= n; ++i) {
            d[i] += d[i - 1];
        }
        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < n; ++i) {
            int j = (s.charAt(i) - 'a' + d[i] % 26 + 26) % 26;
            ans.append((char) ('a' + j));
        }
        return ans.toString();
    }
}

// Accepted solution for LeetCode #2381: Shifting Letters II
func shiftingLetters(s string, shifts [][]int) string {
	n := len(s)
	d := make([]int, n+1)
	for _, e := range shifts {
		if e[2] == 0 {
			e[2]--
		}
		d[e[0]] += e[2]
		d[e[1]+1] -= e[2]
	}
	for i := 1; i <= n; i++ {
		d[i] += d[i-1]
	}
	ans := []byte{}
	for i, c := range s {
		j := (int(c-'a') + d[i]%26 + 26) % 26
		ans = append(ans, byte('a'+j))
	}
	return string(ans)
}

# Accepted solution for LeetCode #2381: Shifting Letters II
class Solution:
    def shiftingLetters(self, s: str, shifts: List[List[int]]) -> str:
        n = len(s)
        d = [0] * (n + 1)
        for i, j, v in shifts:
            if v == 0:
                v = -1
            d[i] += v
            d[j + 1] -= v
        for i in range(1, n + 1):
            d[i] += d[i - 1]
        return ''.join(
            chr(ord('a') + (ord(s[i]) - ord('a') + d[i] + 26) % 26) for i in range(n)
        )

// Accepted solution for LeetCode #2381: Shifting Letters II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2381: Shifting Letters II
// class Solution {
//     public String shiftingLetters(String s, int[][] shifts) {
//         int n = s.length();
//         int[] d = new int[n + 1];
//         for (int[] e : shifts) {
//             if (e[2] == 0) {
//                 e[2]--;
//             }
//             d[e[0]] += e[2];
//             d[e[1] + 1] -= e[2];
//         }
//         for (int i = 1; i <= n; ++i) {
//             d[i] += d[i - 1];
//         }
//         StringBuilder ans = new StringBuilder();
//         for (int i = 0; i < n; ++i) {
//             int j = (s.charAt(i) - 'a' + d[i] % 26 + 26) % 26;
//             ans.append((char) ('a' + j));
//         }
//         return ans.toString();
//     }
// }

// Accepted solution for LeetCode #2381: Shifting Letters II
function shiftingLetters(s: string, shifts: number[][]): string {
    const n: number = s.length;
    const d: number[] = new Array(n + 1).fill(0);

    for (let [i, j, v] of shifts) {
        if (v === 0) {
            v--;
        }
        d[i] += v;
        d[j + 1] -= v;
    }

    for (let i = 1; i <= n; ++i) {
        d[i] += d[i - 1];
    }

    let ans: string = '';
    for (let i = 0; i < n; ++i) {
        const j = (s.charCodeAt(i) - 'a'.charCodeAt(0) + (d[i] % 26) + 26) % 26;
        ans += String.fromCharCode('a'.charCodeAt(0) + j);
    }

    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)

Space

O(1)

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.