Given an array arr[] of N integers, the task is to find the count of subarrays with negative product.
Examples:
Input: arr[] = {-1, 2, -2}
Output: 4
Subarray with negative product are {-1}, {-2}, {-1, 2} and {2, -2}.Input: arr[] = {5, -4, -3, 2, -5}
Output: 8
Approach:
- Replace the positive array elements with 1 and negative array elements with -1.
- Create a prefix product array pre[] where pre[i] stores the product of all the elements from index arr[0] to arr[i].
- Now, it can be noted that the sub-array arr[i…j] has a negative product only if pre[i] * pre[j] is negative.
- Hence, the total count of sub-arrays with negative product will be the product of the count positive and negative elements in the prefix product array.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the count of// subarrays with negative productint negProdSubArr(int arr[], int n){ int positive = 1, negative = 0; for (int i = 0; i < n; i++) { // Replace current element with 1 // if it is positive else replace // it with -1 instead if (arr[i] > 0) arr[i] = 1; else arr[i] = -1; // Take product with previous element // to form the prefix product if (i > 0) arr[i] *= arr[i - 1]; // Count positive and negative elements // in the prefix product array if (arr[i] == 1) positive++; else negative++; } // Return the required count of subarrays return (positive * negative);}// Driver codeint main(){ int arr[] = { 5, -4, -3, 2, -5 }; int n = sizeof(arr) / sizeof(arr[0]); cout << negProdSubArr(arr, n); return (0);} |
Java
// Java implementation of the approach class GFG{ // Function to return the count of // subarrays with negative product static int negProdSubArr(int arr[], int n) { int positive = 1, negative = 0; for (int i = 0; i < n; i++) { // Replace current element with 1 // if it is positive else replace // it with -1 instead if (arr[i] > 0) arr[i] = 1; else arr[i] = -1; // Take product with previous element // to form the prefix product if (i > 0) arr[i] *= arr[i - 1]; // Count positive and negative elements // in the prefix product array if (arr[i] == 1) positive++; else negative++; } // Return the required count of subarrays return (positive * negative); } // Driver code public static void main (String[] args) { int arr[] = { 5, -4, -3, 2, -5 }; int n = arr.length; System.out.println(negProdSubArr(arr, n)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Function to return the count of# subarrays with negative productdef negProdSubArr(arr, n): positive = 1 negative = 0 for i in range(n): # Replace current element with 1 # if it is positive else replace # it with -1 instead if (arr[i] > 0): arr[i] = 1 else: arr[i] = -1 # Take product with previous element # to form the prefix product if (i > 0): arr[i] *= arr[i - 1] # Count positive and negative elements # in the prefix product array if (arr[i] == 1): positive += 1 else: negative += 1 # Return the required count of subarrays return (positive * negative)# Driver codearr = [5, -4, -3, 2, -5]n = len(arr)print(negProdSubArr(arr, n))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System;class GFG{ // Function to return the count of // subarrays with negative product static int negProdSubArr(int []arr, int n) { int positive = 1, negative = 0; for (int i = 0; i < n; i++) { // Replace current element with 1 // if it is positive else replace // it with -1 instead if (arr[i] > 0) arr[i] = 1; else arr[i] = -1; // Take product with previous element // to form the prefix product if (i > 0) arr[i] *= arr[i - 1]; // Count positive and negative elements // in the prefix product array if (arr[i] == 1) positive++; else negative++; } // Return the required count of subarrays return (positive * negative); } // Driver code static public void Main () { int []arr = { 5, -4, -3, 2, -5 }; int n = arr.Length; Console.Write(negProdSubArr(arr, n)); } }// This code is contributed by Sachin. |
Javascript
<script>// Javascript implementation of the approach// Function to return the count of// subarrays with negative productfunction negProdSubArr(arr, n){ let positive = 1, negative = 0; for (let i = 0; i < n; i++) { // Replace current element with 1 // if it is positive else replace // it with -1 instead if (arr[i] > 0) arr[i] = 1; else arr[i] = -1; // Take product with previous element // to form the prefix product if (i > 0) arr[i] *= arr[i - 1]; // Count positive and negative elements // in the prefix product array if (arr[i] == 1) positive++; else negative++; } // Return the required count of subarrays return (positive * negative);}// Driver code let arr = [ 5, -4, -3, 2, -5 ]; let n = arr.length; document.write(negProdSubArr(arr, n));</script> |
Output:
8
Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.
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