Leetcode 1822: Sign of the Product of an Array
You are given an integer array nums. The task is to compute the product of all the elements in the array and return the sign of the product. Implement the function signFunc(x) that returns 1 if x is positive, -1 if x is negative, and 0 if x is 0. Return signFunc(product) for the array.
Problem
Approach
Steps
Complexity
Input: The input consists of an array nums containing integers.
Example: nums = [1, 2, 3, 4, 5]
Constraints:
• 1 <= nums.length <= 1000
• -100 <= nums[i] <= 100
Output: Return the result of signFunc(product) where product is the product of all elements in the array.
Example: Output: 1
Constraints:
• The product can be a very large number, but signFunc only needs to determine the sign of the product.
Goal: Determine the sign of the product of elements in nums.
Steps:
• Initialize a variable to store the product sign (positive, negative, or zero).
• Iterate over the array nums and adjust the sign based on the number of negative elements.
• If any element is zero, return the sign for zero immediately.
• Return the final computed sign.
Goal: Constraints on the input array nums.
Steps:
• 1 <= nums.length <= 1000
• -100 <= nums[i] <= 100
Assumptions:
• The input array is non-empty.
• The product can be large but will not overflow if handled correctly.
• Input: nums = [1, 2, 3, 4, 5]
• Explanation: The product is positive, so the return value is 1.
Approach: The approach is to iterate through the array, multiplying the values and keeping track of whether the product is positive, negative, or zero.
Observations:
• We only need the sign of the product, not the actual product itself.
• If there is a zero in the array, the product is zero, and we can return 0 immediately.
Steps:
• Initialize a variable to keep track of the sign (1 for positive, -1 for negative).
• Iterate through the array: if the number is negative, flip the sign, and if the number is zero, return zero.
• Return the computed sign after finishing the iteration.
Empty Inputs:
• An empty array should not be allowed by the constraints.
Large Inputs:
• Large arrays should be handled efficiently with a time complexity of O(n).
Special Values:
• If the array contains a zero, the result is immediately zero.
Constraints:
• The solution should work for the full range of allowed input sizes and values.
int arraySign(vector<int>& nums) {
int sgn = 1;
for(int i = 0; i < nums.size(); i++) {
if(nums[i] == 0) {
sgn = 0;
break;
} else if(nums[i] > 0) {
} else if(nums[i] < 0) {
sgn *= -1;
}
}
return sgn;
}
1 : Function Definition
int arraySign(vector<int>& nums) {
This is the function definition for `arraySign`, which takes a vector `nums` as input and returns an integer indicating the sign of the product of its elements.
2 : Variable Initialization
int sgn = 1;
The variable `sgn` is initialized to 1, representing the sign of the product. It will be updated during iteration through the array.
3 : Loop Setup
for(int i = 0; i < nums.size(); i++) {
This `for` loop iterates through each element in the `nums` vector to evaluate its effect on the product's sign.
4 : Check for Zero
if(nums[i] == 0) {
This conditional checks if the current element of the array is zero. If found, the product will be zero, so `sgn` is set to 0.
5 : Set Sign to Zero
sgn = 0;
If an element is zero, the sign is immediately set to zero, as multiplying by zero results in zero.
6 : Break Loop
break;
Once a zero is found, the loop is broken, as the sign has already been determined (0).
7 : Check for Positive Number
} else if(nums[i] > 0) {
This checks if the current element is positive. Positive numbers do not change the sign, so nothing needs to be done.
8 : Check for Negative Number
} else if(nums[i] < 0) {
This conditional checks if the current element is negative. Negative numbers flip the sign of the product, so `sgn` is updated accordingly.
9 : Flip the Sign
sgn *= -1;
When a negative number is found, the sign is flipped by multiplying `sgn` by -1.
10 : Return Sign
return sgn;
The final sign of the product is returned. It can be 1 (positive), -1 (negative), or 0 (if any element was zero).
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The solution requires a single pass over the array to compute the result, making it O(n) in time complexity.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) because we only need to keep track of the sign during the iteration.
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