1652.defuse-the-bomb.md

1652. Defuse the Bomb

Brute Force

For each element in the array, we can calculate the sum of the next k elements. If k is negative, we can calculate the sum of the previous k elements.

fun decrypt(code: IntArray, k: Int): IntArray {
    val n = code.size
    val result = IntArray(n)
    if (k == 0) return result

    for (i in code.indices) {
        var sum = 0
        if (k > 0) {
            for (j in i + 1..i + k) {
                sum += code[j % n]
            }
        } else {
            for (j in i - 1 downTo i - (-k)) {
                sum += code[(j + n) % n]
            }
        }
        result[i] = sum
    }
    return result
}

• Time Complexity: O(n * k).
• Space Complexity: O(n).

/**
k = 3
0 1 2 3 4
0 1 2   i
i 1 2 3
  i 2 3 4
0   i 3 4   
0 1   i 4

k = -3
0 1 2 3 4
0 1 2 i
  1 2 3 i
i   2 3 4
0 i   3 4
0 1 i   4
 */

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