LeetCode #1652 — EASY

Defuse the Bomb

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You have a bomb to defuse, and your time is running out! Your informer will provide you with a circular array code of length of n and a key k.

To decrypt the code, you must replace every number. All the numbers are replaced simultaneously.

If k > 0, replace the i^th number with the sum of the next k numbers.
If k < 0, replace the i^th number with the sum of the previous k numbers.
If k == 0, replace the i^th number with 0.

As code is circular, the next element of code[n-1] is code[0], and the previous element of code[0] is code[n-1].

Given the circular array code and an integer key k, return the decrypted code to defuse the bomb!

Example 1:

Input: code = [5,7,1,4], k = 3
Output: [12,10,16,13]
Explanation: Each number is replaced by the sum of the next 3 numbers. The decrypted code is [7+1+4, 1+4+5, 4+5+7, 5+7+1]. Notice that the numbers wrap around.

Example 2:

Input: code = [1,2,3,4], k = 0
Output: [0,0,0,0]
Explanation: When k is zero, the numbers are replaced by 0.

Example 3:

Input: code = [2,4,9,3], k = -2
Output: [12,5,6,13]
Explanation: The decrypted code is [3+9, 2+3, 4+2, 9+4]. Notice that the numbers wrap around again. If k is negative, the sum is of the previous numbers.

Constraints:

n == code.length
1 <= n <= 100
1 <= code[i] <= 100
-(n - 1) <= k <= n - 1

Step 01

Brute Force Baseline

Problem summary: You have a bomb to defuse, and your time is running out! Your informer will provide you with a circular array code of length of n and a key k. To decrypt the code, you must replace every number. All the numbers are replaced simultaneously. If k > 0, replace the ith number with the sum of the next k numbers. If k < 0, replace the ith number with the sum of the previous k numbers. If k == 0, replace the ith number with 0. As code is circular, the next element of code[n-1] is code[0], and the previous element of code[0] is code[n-1]. Given the circular array code and an integer key k, return the decrypted code to defuse the bomb!

Baseline thinking

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

Pattern signal: Array · Sliding Window

Example 1

[5,7,1,4]
3

Example 2

[1,2,3,4]
0

Example 3

[2,4,9,3]
-2

Core Insight

What unlocks the optimal approach

As the array is circular, use modulo to find the correct index.
The constraints are low enough for a brute-force solution.

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 #1652: Defuse the Bomb
class Solution {
    public int[] decrypt(int[] code, int k) {
        int n = code.length;
        int[] ans = new int[n];
        if (k == 0) {
            return ans;
        }
        for (int i = 0; i < n; ++i) {
            if (k > 0) {
                for (int j = i + 1; j < i + k + 1; ++j) {
                    ans[i] += code[j % n];
                }
            } else {
                for (int j = i + k; j < i; ++j) {
                    ans[i] += code[(j + n) % n];
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1652: Defuse the Bomb
func decrypt(code []int, k int) []int {
	n := len(code)
	ans := make([]int, n)
	if k == 0 {
		return ans
	}
	for i := 0; i < n; i++ {
		if k > 0 {
			for j := i + 1; j < i+k+1; j++ {
				ans[i] += code[j%n]
			}
		} else {
			for j := i + k; j < i; j++ {
				ans[i] += code[(j+n)%n]
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #1652: Defuse the Bomb
class Solution:
    def decrypt(self, code: List[int], k: int) -> List[int]:
        n = len(code)
        ans = [0] * n
        if k == 0:
            return ans
        for i in range(n):
            if k > 0:
                for j in range(i + 1, i + k + 1):
                    ans[i] += code[j % n]
            else:
                for j in range(i + k, i):
                    ans[i] += code[(j + n) % n]
        return ans

// Accepted solution for LeetCode #1652: Defuse the Bomb
struct Solution;

impl Solution {
    fn decrypt(code: Vec<i32>, k: i32) -> Vec<i32> {
        let n = code.len();
        if k == 0 {
            return vec![0; n];
        }
        let mut res = vec![];
        for i in 0..n {
            let mut sum = 0;
            for j in 0..k.abs() {
                let index = if k > 0 {
                    (n as i32 + (i as i32 + 1 + j)) as usize % n
                } else {
                    (n as i32 + (i as i32 - 1 - j)) as usize % n
                };
                sum += code[index];
            }
            res.push(sum);
        }
        res
    }
}
#[test]
fn test() {
    let code = vec![5, 7, 1, 4];
    let k = 3;
    let res = vec![12, 10, 16, 13];
    assert_eq!(Solution::decrypt(code, k), res);
    let code = vec![1, 2, 3, 4];
    let k = 0;
    let res = vec![0, 0, 0, 0];
    assert_eq!(Solution::decrypt(code, k), res);
    let code = vec![2, 4, 9, 3];
    let k = -2;
    let res = vec![12, 5, 6, 13];
    assert_eq!(Solution::decrypt(code, k), res);
}

// Accepted solution for LeetCode #1652: Defuse the Bomb
function decrypt(code: number[], k: number): number[] {
    const n: number = code.length;
    const ans: number[] = Array(n).fill(0);

    if (k === 0) {
        return ans;
    }

    for (let i = 0; i < n; ++i) {
        if (k > 0) {
            for (let j = i + 1; j < i + k + 1; ++j) {
                ans[i] += code[j % n];
            }
        } else {
            for (let j = i + k; j < i; ++j) {
                ans[i] += code[(j + n) % n];
            }
        }
    }

    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 × |k|)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

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.

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.