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Leetcode 1652. Defuse the Bomb

Problem Statement

You have a bomb to defuse, and your experience tells you that defusing it will be much easier once you understand the code used to create this bomb.

You are given an integer array code and an integer k. The array code is of length n and is indexed from 0 to n - 1. The rounding rule for calculating the array value at index i is as follows:

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].

Return the transformed array code into the new array as described above.

Clarifying Questions

  1. Can the array contain negative numbers?
    • Yes, the array can contain any integers.
  2. What is the range of values for k?
    • k can be any integer between -n and n inclusive.
  3. What constraints do we have on the length of the array n?
    • The length of the array n will be between 1 and 10000.

Strategy

  1. Initial Checks:
    • If k == 0, return an array of zeroes of the same length as code.
  2. Create the Circular Array:
    • Since the array is circular, we need to handle wrap-around indices properly. This can be managed by treating the array as circular using modulo arithmetic.
  3. Iterate and Compute:
    • For each element in the array, compute the sum of the next/previous k elements based on the value of k.
    • Use a sliding window approach where we keep a sum of the current window and adjust it as we slide through the array.

Time Complexity

Code

#include <vector>
using namespace std;

vector<int> decrypt(vector<int>& code, int k) {
    int n = code.size();
    vector<int> result(n, 0);
    
    if (k == 0) {
        return result;
    }
    
    int start = k > 0 ? 1 : k;
    int end = k > 0 ? k : -1;
    int current_sum = 0;
    
    // Initial window sum calculation
    for (int i = start; i <= end; ++i) {
        current_sum += code[(i + n) % n];
    }
    
    for (int i = 0; i < n; ++i) {
        result[i] = current_sum;
        
        // Slide window right
        current_sum -= code[(i + start + n) % n];
        current_sum += code[(i + end + 1 + n) % n];
    }
    
    return result;
}

Explanation of the Code

  1. Initialization:
    • Create a result vector initialized to zero.
    • Handle the special case where k == 0.
  2. Sliding Window Setup:
    • Depending on whether k is positive or negative, set up the initial sum of the window.
    • Use modulo arithmetic to handle the circular nature of the array.
  3. Sliding the Window:
    • For each position in the array, update the result with the current sum.
    • Slide the window by subtracting the element that is out of the window and adding the next element to the window.
    • Use modulo to ensure indices stay within bounds.

By following this approach, we ensure that the array is correctly transformed as per the problem’s requirements in an efficient manner.

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