Leetcode problem: 1652. Defuse the Bomb

Description:

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!

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

Solution:

class Solution {
    public int[] decrypt(int[] code, int k) {
        int len = code.length;
        int[] result = new int[len];
        int sum;

        for(int i=0;i<len;i++){
            if(k>0){
              sum= getSum(code,i+1,k,len,"right");
              result[i] = sum;
            }
            else if(k<0){
              sum= getSum(code,i-1,Math.abs(k),len,"left");
              result[i] = sum;
            }
            else{
              result[i] = 0;
            }
              
        }
        return result;
    }

    int getSum(int[] code, int i, int k, int len, String direction){
        int sum = 0;
        int index;
        if(direction.equals("right")){
            for(int j=i;j<i+k;j++){
              if(j>len-1){
                index = j-len;
              }
              else{
                index = j;
              }
              sum+=code[index];
            }
        }
        else{
            for(int j=i;j>i-k;j--){
              if(j<0){
                index = j+len;
              }
              else{
                index = j;
              }
              sum+=code[index];
            }
        }
        return sum;
    }
}

Approach:

The goal of this algorithm is to transform the input array code into a new array result, based on the integer value k. The transformation depends on whether k is positive, negative, or zero.

1. Initialization

• First, we create a new array named result that has the same length as the input array code.
• This array will store the computed values for each index after applying the transformation rule.
• Initially, all elements in result can be set to 0.

result = [0] * len(code)

This ensures we have a placeholder for every element we’ll later update.

2. Iterating Through Each Element

We loop through each index i of the input array code.
For every position i, we will determine which elements to sum based on the value of k.

3. Case 1: When k > 0

• This means for every element at position i, we need to compute the sum of the next k elements in the array.
• The array is circular, so if we reach the end of the array, we wrap around to the beginning.

Example:
If code = [5, 7, 1, 4] and k = 2:

• For index 0: next two elements are [7, 1], sum = 8
• For index 1: next two elements are [1, 4], sum = 5
• For index 2: next two elements are [4, 5] (wraps around), sum = 9
• For index 3: next two elements are [5, 7], sum = 12

To achieve wrapping, we use modular arithmetic (% n), where n is the length of code.

result[i] = sum(code[(i + j) % n] for j in range(1, k + 1))

4. Case 2: When k < 0

• This means we need to compute the sum of the previous |k| elements for each index.
• Again, since the array is circular, we can wrap around to the end of the array when needed.

Example:
If code = [5, 7, 1, 4] and k = -2:

• For index 0: previous two elements are [4, 1], sum = 5
• For index 1: previous two elements are [5, 4], sum = 9
• For index 2: previous two elements are [7, 5], sum = 12
• For index 3: previous two elements are [1, 7], sum = 8

This can also be implemented with modular arithmetic:

result[i] = sum(code[(i + j) % n] for j in range(k, 0))

Here, range(k, 0) generates negative offsets that move backward from index i.

5. Case 3: When k == 0

• If k is zero, it means no elements before or after the current position should be considered.
• Therefore, every element in result becomes zero.

if k == 0:
    return [0] * len(code)

6. Final Output

After processing all elements, the result array contains the transformed values according to the rules defined by k.

✅ Example Summary