LeetCode #3361 — MEDIUM

Shift Distance Between Two Strings

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

Solve on LeetCode

The Problem

Problem Statement

You are given two strings s and t of the same length, and two integer arrays nextCost and previousCost.

In one operation, you can pick any index i of s, and perform either one of the following actions:

Shift s[i] to the next letter in the alphabet. If s[i] == 'z', you should replace it with 'a'. This operation costs nextCost[j] where j is the index of s[i] in the alphabet.
Shift s[i] to the previous letter in the alphabet. If s[i] == 'a', you should replace it with 'z'. This operation costs previousCost[j] where j is the index of s[i] in the alphabet.

The shift distance is the minimum total cost of operations required to transform s into t.

Return the shift distance from s to t.

Example 1:

Input: s = "abab", t = "baba", nextCost = [100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], previousCost = [1,100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

Output: 2

Explanation:

We choose index i = 0 and shift s[0] 25 times to the previous character for a total cost of 1.
We choose index i = 1 and shift s[1] 25 times to the next character for a total cost of 0.
We choose index i = 2 and shift s[2] 25 times to the previous character for a total cost of 1.
We choose index i = 3 and shift s[3] 25 times to the next character for a total cost of 0.

Example 2:

Input: s = "leet", t = "code", nextCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], previousCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]

Output: 31

Explanation:

We choose index i = 0 and shift s[0] 9 times to the previous character for a total cost of 9.
We choose index i = 1 and shift s[1] 10 times to the next character for a total cost of 10.
We choose index i = 2 and shift s[2] 1 time to the previous character for a total cost of 1.
We choose index i = 3 and shift s[3] 11 times to the next character for a total cost of 11.

Constraints:

1 <= s.length == t.length <= 10⁵
s and t consist only of lowercase English letters.
nextCost.length == previousCost.length == 26
0 <= nextCost[i], previousCost[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given two strings s and t of the same length, and two integer arrays nextCost and previousCost. In one operation, you can pick any index i of s, and perform either one of the following actions: Shift s[i] to the next letter in the alphabet. If s[i] == 'z', you should replace it with 'a'. This operation costs nextCost[j] where j is the index of s[i] in the alphabet. Shift s[i] to the previous letter in the alphabet. If s[i] == 'a', you should replace it with 'z'. This operation costs previousCost[j] where j is the index of s[i] in the alphabet. The shift distance is the minimum total cost of operations required to transform s into t. Return the shift distance from s to t.

Baseline thinking

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

Pattern signal: Array

Example 1

"abab"
"baba"
[100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
[1,100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

Example 2

"leet"
"code"
[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]
[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]

Core Insight

What unlocks the optimal approach

- For every unordered pair of characters <code>(a, b)</code>, the cost of turning <code>a</code> into <code>b</code> is equal to the minimum between: <ul> <li>If <code>i < j</code>, <code>nextCost[i] + nextCost[i + 1] + … + nextCost[j - 1]</code>, and <code>nextCost[i] + nextCost[i + 1] + … + nextCost[25] + nextCost[0] + … + nextCost[j - 1]</code> otherwise.</li> <li>If <code>i < j</code>, <code>prevCost[i] + prevCost[i - 1] + … + prevCost[0] + prevCost[25] + … + prevCost[j + 1]</code>, and <code>prevCost[i] + prevCost[i - 1] + … + prevCost[j + 1]</code> otherwise.</li> </ul> Where <code>i</code> and <code>j</code> are the indices of <code>a</code> and <code>b</code> in the alphabet.
The shift distance is the sum of costs of turning <code>s[i]</code> into <code>t[i]</code>.

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 #3361: Shift Distance Between Two Strings
class Solution {
    public long shiftDistance(String s, String t, int[] nextCost, int[] previousCost) {
        int m = 26;
        long[] s1 = new long[(m << 1) + 1];
        long[] s2 = new long[(m << 1) + 1];
        for (int i = 0; i < (m << 1); i++) {
            s1[i + 1] = s1[i] + nextCost[i % m];
            s2[i + 1] = s2[i] + previousCost[(i + 1) % m];
        }
        long ans = 0;
        for (int i = 0; i < s.length(); i++) {
            int x = s.charAt(i) - 'a';
            int y = t.charAt(i) - 'a';
            long c1 = s1[y + (y < x ? m : 0)] - s1[x];
            long c2 = s2[x + (x < y ? m : 0)] - s2[y];
            ans += Math.min(c1, c2);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3361: Shift Distance Between Two Strings
func shiftDistance(s string, t string, nextCost []int, previousCost []int) (ans int64) {
	m := 26
	s1 := make([]int64, (m<<1)+1)
	s2 := make([]int64, (m<<1)+1)
	for i := 0; i < (m << 1); i++ {
		s1[i+1] = s1[i] + int64(nextCost[i%m])
		s2[i+1] = s2[i] + int64(previousCost[(i+1)%m])
	}
	for i := 0; i < len(s); i++ {
		x := int(s[i] - 'a')
		y := int(t[i] - 'a')
		z := y
		if y < x {
			z += m
		}
		c1 := s1[z] - s1[x]
		z = x
		if x < y {
			z += m
		}
		c2 := s2[z] - s2[y]
		ans += min(c1, c2)
	}
	return
}

# Accepted solution for LeetCode #3361: Shift Distance Between Two Strings
class Solution:
    def shiftDistance(
        self, s: str, t: str, nextCost: List[int], previousCost: List[int]
    ) -> int:
        m = 26
        s1 = [0] * (m << 1 | 1)
        s2 = [0] * (m << 1 | 1)
        for i in range(m << 1):
            s1[i + 1] = s1[i] + nextCost[i % m]
            s2[i + 1] = s2[i] + previousCost[(i + 1) % m]
        ans = 0
        for a, b in zip(s, t):
            x, y = ord(a) - ord("a"), ord(b) - ord("a")
            c1 = s1[y + m if y < x else y] - s1[x]
            c2 = s2[x + m if x < y else x] - s2[y]
            ans += min(c1, c2)
        return ans

// Accepted solution for LeetCode #3361: Shift Distance Between Two Strings
// 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 #3361: Shift Distance Between Two Strings
// class Solution {
//     public long shiftDistance(String s, String t, int[] nextCost, int[] previousCost) {
//         int m = 26;
//         long[] s1 = new long[(m << 1) + 1];
//         long[] s2 = new long[(m << 1) + 1];
//         for (int i = 0; i < (m << 1); i++) {
//             s1[i + 1] = s1[i] + nextCost[i % m];
//             s2[i + 1] = s2[i] + previousCost[(i + 1) % m];
//         }
//         long ans = 0;
//         for (int i = 0; i < s.length(); i++) {
//             int x = s.charAt(i) - 'a';
//             int y = t.charAt(i) - 'a';
//             long c1 = s1[y + (y < x ? m : 0)] - s1[x];
//             long c2 = s2[x + (x < y ? m : 0)] - s2[y];
//             ans += Math.min(c1, c2);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3361: Shift Distance Between Two Strings
function shiftDistance(s: string, t: string, nextCost: number[], previousCost: number[]): number {
    const m = 26;
    const s1: number[] = Array((m << 1) + 1).fill(0);
    const s2: number[] = Array((m << 1) + 1).fill(0);
    for (let i = 0; i < m << 1; i++) {
        s1[i + 1] = s1[i] + nextCost[i % m];
        s2[i + 1] = s2[i] + previousCost[(i + 1) % m];
    }
    let ans = 0;
    const a = 'a'.charCodeAt(0);
    for (let i = 0; i < s.length; i++) {
        const x = s.charCodeAt(i) - a;
        const y = t.charCodeAt(i) - a;
        const c1 = s1[y + (y < x ? m : 0)] - s1[x];
        const c2 = s2[x + (x < y ? m : 0)] - s2[y];
        ans += Math.min(c1, c2);
    }
    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.