grid47 Exploring patterns and algorithms
Sep 22, 2024
7 min read
Solution to LeetCode 457: Circular Array Loop Problem
You are playing a game with a circular array where each element indicates how many steps to move forward or backward. Determine if there exists a cycle in the array where the cycle has more than one element and all elements in the cycle move in the same direction. Return true if such a cycle exists, otherwise return false.
Problem
Approach
Steps
Complexity
Input: The input is a circular array of non-zero integers where each element specifies the number of steps to move forward or backward.
Example: nums = [2, -1, 1, 2, 2]
Constraints:
• 1 <= nums.length <= 5000
• -1000 <= nums[i] <= 1000
• nums[i] != 0
Output: Return a boolean indicating whether there exists a valid cycle in the array.
Example: Output: true
Constraints:
• The output should be a boolean indicating whether a valid cycle exists.
Goal: The goal is to identify if there is a cycle in the array that satisfies the conditions outlined above.
Steps:
• 1. Start iterating over the array, marking visited elements.
• 2. Follow the directions specified by the values at each index, checking for a cycle.
• 3. Ensure that the cycle formed involves moving entirely in one direction (either all forward or all backward).
Goal: Ensure that the solution works efficiently for large inputs while maintaining constant extra space.
Steps:
• The array length can be up to 5000.
• Each value in the array can range from -1000 to 1000.
Assumptions:
• The input array is non-empty.
• Input: nums = [2, -1, 1, 2, 2]
• Explanation: There is a valid cycle starting at index 0, which leads to indices 2, 3, and back to 0, forming a valid cycle.
• Input: nums = [-1, -2, -3, -4, -5, 6]
• Explanation: There are no valid cycles as the only cycle is of size 1, which is not valid.
Approach: The solution involves tracking each element's direction and checking if it forms a valid cycle according to the problem's requirements.
Observations:
• The array is circular, so each element can wrap around to the start or end.
• A cycle can be detected by tracking movements and ensuring the same direction for all cycle elements.
Steps:
• 1. Start iterating over the array and mark each element as visited.
• 2. For each unvisited element, track its movement and direction.
• 3. If the movement leads back to the starting index with the same direction, return true.
• 4. If no cycle is found, return false.
Empty Inputs:
• If the array has only one element, return false.
Large Inputs:
• Ensure that the solution works efficiently with input arrays up to the maximum size of 5000.
Special Values:
• If the array contains alternating positive and negative values, ensure that no cycle is falsely identified.
Constraints:
• Handle large inputs efficiently, ensuring an O(n) time complexity solution with constant space.
boolcircularArrayLoop(vector<int>& nums) {
int n = nums.size();
if(n ==1) returnfalse;
for(int i =0; i < n; i++) {
if(nums[i] ==0) continue;
int j = i, k = getIdx(i, nums);
while(nums[k] * nums[i] >0&& nums[getIdx(k, nums)] * nums[i] >0) {
if(j == k) {
if(j == getIdx(j, nums)) break;
returntrue;
}
j = getIdx(j, nums);
k = getIdx(getIdx(k, nums), nums);
}
j = i;
int val = nums[i];
while(nums[j] * val >0) {
int next = getIdx(j, nums);
nums[j] =0;
j = next;
}
}
returnfalse;
}
intgetIdx(int i, vector<int>&nums) {
int n = nums.size();
return i + nums[i] >=0? (i + nums[i]) %n: n + ((i + nums[i]) % n);
}
1 : Function Definition
boolcircularArrayLoop(vector<int>& nums) {
Defines the `circularArrayLoop` function, which takes a vector `nums` and returns a boolean indicating whether a cycle exists in the array.
2 : Variable Initialization
int n = nums.size();
Initializes variable `n` to store the size of the input array `nums`.
3 : Edge Case Handling
if(n ==1) returnfalse;
Checks if the size of the array is 1. A single element cannot form a cycle, so the function returns `false`.
4 : Loop Initialization
for(int i =0; i < n; i++) {
Starts a loop that iterates over each element in the array `nums`.
5 : Edge Case Handling
if(nums[i] ==0) continue;
Skips the current element if it is 0, as it cannot contribute to a valid cycle.
6 : Variable Initialization
int j = i, k = getIdx(i, nums);
Initializes two pointers: `j` (the slow pointer) starting at index `i`, and `k` (the fast pointer) calculated using the `getIdx` helper function.
Enters a while loop that runs as long as the elements at the slow and fast pointers (`nums[i]`, `nums[k]`) have the same sign (either both positive or both negative), meaning they are moving in the same direction.
8 : Cycle Detection
if(j == k) {
Checks if the slow pointer (`j`) and the fast pointer (`k`) meet, indicating a potential cycle.
9 : Cycle Detection
if(j == getIdx(j, nums)) break;
If the slow pointer meets itself (i.e., points to its own index), the loop breaks because it’s not part of a valid cycle.
10 : Cycle Detection
returntrue;
If the slow pointer meets the fast pointer and the cycle is valid, the function returns `true`, indicating a cycle has been found.
11 : Pointer Update
j = getIdx(j, nums);
Moves the slow pointer (`j`) to the next index calculated by the `getIdx` helper function.
12 : Pointer Update
k = getIdx(getIdx(k, nums), nums);
Moves the fast pointer (`k`) to the next index, jumping two steps ahead by calling `getIdx` twice.
13 : Pointer Reset
j = i;
Resets the slow pointer (`j`) back to the current index `i`.
14 : Variable Initialization
int val = nums[i];
Stores the value at `nums[i]` in the variable `val` to help mark the elements as visited.
15 : Marking Visited
while(nums[j] * val >0) {
Enters a while loop to mark all elements in the current cycle as visited (setting them to 0), preventing revisiting them in future iterations.
16 : Pointer Update
int next = getIdx(j, nums);
Calculates the next index in the cycle using the `getIdx` helper function.
17 : Marking Visited
nums[j] =0;
Marks the current element as visited by setting it to 0.
18 : Pointer Update
j = next;
Moves the slow pointer (`j`) to the next index in the cycle.
19 : Return Statement
returnfalse;
Returns `false` if no cycle was detected in the array.
20 : Helper Function
intgetIdx(int i, vector<int>&nums) {
Defines the helper function `getIdx`, which calculates the next index in the circular array using modulo arithmetic.
21 : Helper Function
int n = nums.size();
Gets the size of the array `nums`.
22 : Helper Function
return i + nums[i] >=0? (i + nums[i]) %n: n + ((i + nums[i]) % n);
Returns the next index after applying the circular wrap-around logic using modulo arithmetic.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) since we iterate through the array once, checking for cycles.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) as no extra space is required apart from a few variables.
