Leetcode 1652: Defuse the Bomb
You are given a circular array ‘code’ and an integer ‘k’. To decrypt the code, you must replace every element in the array based on the value of k. If k > 0, replace the i-th element with the sum of the next k elements. If k < 0, replace the i-th element with the sum of the previous k elements. If k == 0, replace the i-th element with 0. The array is circular, meaning the next element of the last one is the first, and the previous element of the first one is the last.
Problem
Approach
Steps
Complexity
Input: An array of integers representing the encrypted code and an integer representing the decryption key k.
Example: code = [4, 6, 3, 7], k = 2
Constraints:
• 1 <= n <= 100
• 1 <= code[i] <= 100
• -(n - 1) <= k <= n - 1
Output: Return an array where each element is replaced according to the decryption rules based on k.
Example: Output: [13, 11, 13, 13]
Constraints:
Goal: To efficiently replace each element in the array based on the value of k, considering the circular nature of the array.
Steps:
• Check if k is zero, if so return an array of zeros.
• If k > 0, for each element in the array, sum the next k elements (consider the circular nature).
• If k < 0, for each element in the array, sum the previous k elements (consider the circular nature).
• Return the resulting array after all replacements.
Goal: The problem must handle various input sizes and values for code and k.
Steps:
• 1 <= n <= 100
• 1 <= code[i] <= 100
• -(n - 1) <= k <= n - 1
Assumptions:
• The input will always contain valid values for code and k.
• Input: code = [4, 6, 3, 7], k = 2
• Explanation: We sum the next k elements for each element in the array, taking care of the circular behavior where elements wrap around.
Approach: The approach involves iterating through the array and applying the sum rule based on k for each element, handling the circular nature of the array with modular arithmetic.
Observations:
• We need to consider both positive and negative values of k, and ensure that the array is treated as circular.
• The modulo operator can help in wrapping the indices around when summing elements beyond the boundaries of the array.
Steps:
• If k is 0, return an array of zeros.
• For k > 0, start from the first element and sum the next k elements, using modulo to handle the circular array.
• For k < 0, sum the previous k elements in the same way.
• Return the modified array.
Empty Inputs:
• Ensure the solution handles arrays with one element.
Large Inputs:
• Ensure the solution works within the constraints when n is at its maximum value.
Special Values:
• Handle the case where k == 0 by returning an array of zeros.
Constraints:
• The array must always have a size between 1 and 100.
vector<int> decrypt(vector<int>& code, int k) {
int n = code.size();
vector<int> ans(n, 0);
if(k == 0) return ans;
int sum = 0;
if(k > 0) {
for(int i = 1; i < k + 1; i++)
sum += code[i];
ans[0] = sum;
for(int i = 1; i < n; i++) {
sum += code[(i + k) % n];
sum -= code[i];
ans[i] = sum;
}
return ans;
}
for(int i = 0; i < abs(k); i++)
sum += code[n - 1 - i];
ans[0] = sum;
for(int i = 1; i < n; i++) {
sum += code[i - 1];
sum -= code[(i - 1 - abs(k) + n) % n];
ans[i] = sum;
}
return ans;
}
1 : Function Declaration
vector<int> decrypt(vector<int>& code, int k) {
Declares the decrypt function that takes an array and an integer k.
2 : Variable Initialization
int n = code.size();
Calculates the size of the input array.
3 : Variable Initialization
vector<int> ans(n, 0);
Initializes the answer array with zeros.
4 : Conditional Statement
if(k == 0) return ans;
Handles the special case where k is zero, returning a zero-filled array.
5 : Variable Initialization
int sum = 0;
Initializes a variable to store the sum of elements.
6 : Conditional Statement
if(k > 0) {
Begins processing for positive values of k.
7 : Loop
for(int i = 1; i < k + 1; i++)
Iterates over the first k elements after the current element.
8 : Sum Calculation
sum += code[i];
Adds the value of the current element to the sum.
9 : Array Assignment
ans[0] = sum;
Stores the initial sum in the answer array.
10 : Loop
for(int i = 1; i < n; i++) {
Iterates through the array to compute sums for each position.
11 : Sum Adjustment
sum += code[(i + k) % n];
Adds the next element in the k range to the sum.
12 : Sum Adjustment
sum -= code[i];
Removes the element that goes out of the k range from the sum.
13 : Array Assignment
ans[i] = sum;
Stores the computed sum for the current position in the answer array.
14 : Return Statement
return ans;
Returns the computed answer array.
15 : Loop
for(int i = 0; i < abs(k); i++)
Iterates over the last k elements for negative k.
16 : Sum Calculation
sum += code[n - 1 - i];
Adds the value of the current element to the sum for negative k.
17 : Array Assignment
ans[0] = sum;
Stores the initial sum for negative k in the answer array.
18 : Loop
for(int i = 1; i < n; i++) {
Iterates through the array for sums with negative k.
19 : Sum Adjustment
sum += code[i - 1];
Adds the next element to the sum for negative k.
20 : Sum Adjustment
sum -= code[(i - 1 - abs(k) + n) % n];
Removes the element going out of the range for negative k.
21 : Array Assignment
ans[i] = sum;
Stores the computed sum for the current position in the answer array.
22 : Return Statement
return ans;
Returns the computed answer array for negative k.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) because we iterate over the array once.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the array we create to store the result.
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