Leetcode 2529: Maximum Count of Positive Integer and Negative Integer
You are given an array of integers sorted in non-decreasing order. Your task is to determine the maximum count between the number of positive integers and the number of negative integers in the array. Zero is not considered positive or negative.
Problem
Approach
Steps
Complexity
Input: The input consists of a list of integers `nums` which is sorted in non-decreasing order.
Example: [-3, -2, -1, 1, 2, 3]
Constraints:
• 1 <= nums.length <= 2000
• -2000 <= nums[i] <= 2000
• nums is sorted in non-decreasing order.
Output: Return the maximum of the count of positive integers and the count of negative integers in the array.
Example: 3
Constraints:
• The output will be a single integer representing the maximum count of positive or negative integers.
Goal: To determine the count of positive and negative integers in the array and return the maximum of the two.
Steps:
• Initialize counters for positive and negative integers.
• Iterate through the array and count the positive and negative integers.
• Return the maximum of the two counts.
Goal: The array will have a length of at least 1, and the integers will be in the range of -2000 to 2000. The array is sorted in non-decreasing order.
Steps:
• 1 <= nums.length <= 2000
• -2000 <= nums[i] <= 2000
Assumptions:
• The input array is sorted in non-decreasing order.
• Zero (0) is neither positive nor negative.
• Input: [-3, -2, -1, 1, 2, 3]
• Explanation: In this example, there are 3 positive integers and 3 negative integers. The maximum count between them is 3.
• Input: [-5, -3, -1, 0, 2, 4, 6]
• Explanation: In this example, there are 3 positive integers and 3 negative integers. The maximum count between them is 4.
• Input: [0, 2, 4, 8, 10]
• Explanation: In this example, there are 5 positive integers and 0 negative integers. The maximum count between them is 5.
Approach: The solution involves counting the number of positive and negative integers in the array and returning the maximum of the two. The input is already sorted, which helps us efficiently determine the counts.
Observations:
• The array is sorted, which means the negative integers will be at the beginning and positive integers will be at the end.
• Counting positive and negative integers separately is the most straightforward approach.
• The problem could potentially be optimized further if we use binary search to quickly find the boundary between negative and positive numbers, reducing the time complexity.
Steps:
• Initialize two counters for positive and negative integers.
• Iterate through the array and update the counters based on the value of each integer.
• Return the maximum of the two counters.
Empty Inputs:
• The input array will always have at least one element, as per the constraints.
Large Inputs:
• Ensure that the solution efficiently handles arrays of size up to 2000 elements.
Special Values:
• Arrays with zero values should be handled correctly, as zero is neither positive nor negative.
Constraints:
• The array is sorted, so you can exploit this property for more efficient solutions.
int maximumCount(vector<int>& nums) {
int pos = 0, neg = 0;
for(int i = 0; i < nums.size(); i++)
if(nums[i] > 0) pos++;
else if(nums[i] < 0) neg++;
return max(pos, neg);
}
1 : Function Definition
int maximumCount(vector<int>& nums) {
This is the function definition for 'maximumCount', which takes a reference to a vector of integers 'nums' and returns an integer value.
2 : Variable Initialization
int pos = 0, neg = 0;
Two integer variables, 'pos' and 'neg', are initialized to 0. 'pos' will track the count of positive numbers, and 'neg' will track the count of negative numbers.
3 : Loop
for(int i = 0; i < nums.size(); i++)
This for loop iterates through each element in the 'nums' vector, with 'i' as the loop index.
4 : Conditional Statement
if(nums[i] > 0) pos++;
If the current element is positive, the 'pos' counter is incremented.
5 : Conditional Statement
else if(nums[i] < 0) neg++;
If the current element is negative, the 'neg' counter is incremented.
6 : Return Statement
return max(pos, neg);
After completing the loop, the function returns the maximum of 'pos' and 'neg', which represents the greater count between positive and negative numbers.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) because we need to count the number of positive and negative integers by iterating through the entire array.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) because we only need a constant amount of space for the counters.
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