Minimum operations to make product of adjacent element pair of prefix sum negative
Given an array arr[ ] of size N, consider an array prefix[ ] where prefix[i] is the sum of the first i elements of arr. The task is to find the minimum number of operations required to modify the given array such that the product of any two adjacent elements in the prefix array is negative. In one operation you can increment or decrement the value of any element by 1.
Input: arr[] = {1, -3, 1, 0}
Output: 4
Explanation: The sequence can be transformed into 1, -2, 2, -2 by 4 operations, the sum of the prefixes are 1, -1, 1, -1 which satisfy all the conditions.Input: arr[] = {-1 4 3 2 -5 4}
Output: 8
Approach: The idea is to try out two independent possibilities that either even length prefix sums are positive and odd length prefix sums are negative or vice versa. Follow the steps below to solve the problem:
- Initialize a variable res = INT_MAX to store the minimum number of operations.
- Traverse over the range [0, 1] using the variable r.
- Initialize variables ans = 0 and sum = 0 to store the count of total operations and the current prefix sum respectively.
- Traverse over the range [0, N-1] using the variable i,
- Add arr[i] to the value of sum.
- If the value of (i+r) is odd and if sum is not positive then add sum+1 to ans and update the value of sum to 1.
- Otherwise if (i+r) is even and if sum is not negative then add sum+1 to ans update the value of sum to -1.
- Update the value of res as min(res, ans).
- After completing the above steps print the value of res.
C++
// C++ code for the above approach #include <bits/stdc++.h> using namespace std; // Function to find minimum operations // needed to make the product of any // two adjacent elements in prefix // sum array negative void minOperations(vector< int > a) { // Stores the minimum operations int res = INT_MAX; int N = a.size(); for ( int r = 0; r < 2; r++) { // Stores the prefix sum // and number of operations int sum = 0, ans = 0; // Traverse the array for ( int i = 0; i < N; i++) { // Update the value of sum sum += a[i]; // Check if i+r is odd if ((i + r) % 2) { // Check if prefix sum // is not positive if (sum <= 0) { // Update the value of // ans and sum ans += -sum + 1; sum = 1; } } else { // Check if prefix sum is // not negative if (sum >= 0) { // Update the value of // ans and sum ans += sum + 1; sum = -1; } } } // Update the value of res res = min(res, ans); } // Print the value of res cout << res; } // Driver Code int main() { vector< int > a{ 1, -3, 1, 0 }; minOperations(a); return 0; } |
Java
// Java code for the above approach import java.util.*; class GFG{ // Function to find minimum operations // needed to make the product of any // two adjacent elements in prefix // sum array negative static void minOperations(ArrayList<Integer> a) { // Stores the minimum operations int res = Integer.MAX_VALUE; int N = a.size(); for ( int r = 0 ; r < 2 ; r++) { // Stores the prefix sum // and number of operations int sum = 0 , ans = 0 ; // Traverse the array for ( int i = 0 ; i < N; i++) { // Update the value of sum sum += a.get(i); // Check if i+r is odd if ((i + r) % 2 == 1 ) { // Check if prefix sum // is not positive if (sum <= 0 ) { // Update the value of // ans and sum ans += -sum + 1 ; sum = 1 ; } } else { // Check if prefix sum is // not negative if (sum >= 0 ) { // Update the value of // ans and sum ans += sum + 1 ; sum = - 1 ; } } } // Update the value of res res = Math.min(res, ans); } // Print the value of res System.out.print(res); } // Driver Code public static void main(String args[]) { ArrayList<Integer> a = new ArrayList<Integer>(); a.add( 1 ); a.add(- 3 ); a.add( 1 ); a.add( 0 ); minOperations(a); } } // This code is contributed by SURENDRA_GANGWAR |
Python3
# python code for the above approach # // Function to find minimum operations # // needed to make the product of any # // two adjacent elements in prefix # // sum array negative def minOperations(a): #Stores the minimum operations res = 100000000000 N = len (a) for r in range ( 0 , 2 ): # Stores the prefix sum # and number of operations sum = 0 ans = 0 # Traverse the array for i in range ( 0 ,N): # Update the value of sum sum + = a[i] # Check if i+r is odd if ((i + r) % 2 ): # Check if prefix sum # is not positive if ( sum < = 0 ): # Update the value of # ans and sum ans + = - sum + 1 sum = 1 else : # Check if prefix sum is # not negative if ( sum > = 0 ): # Update the value of # ans and sum ans + = sum + 1 ; sum = - 1 ; # Update the value of res res = min (res, ans) # // Print the value of res print (res) # Driver code a = [ 1 , - 3 , 1 , 0 ] minOperations(a); # This code is contributed by Stream_Cipher |
C#
// C# code for the above approach using System; using System.Collections.Generic; class GFG { // Function to find minimum operations // needed to make the product of any // two adjacent elements in prefix // sum array negative static void minOperations(List< int > a) { // Stores the minimum operations int res = Int32.MaxValue; int N = a.Count; for ( int r = 0; r < 2; r++) { // Stores the prefix sum // and number of operations int sum = 0, ans = 0; // Traverse the array for ( int i = 0; i < N; i++) { // Update the value of sum sum += a[i]; // Check if i+r is odd if ((i + r) % 2 == 1) { // Check if prefix sum // is not positive if (sum <= 0) { // Update the value of // ans and sum ans += -sum + 1; sum = 1; } } else { // Check if prefix sum is // not negative if (sum >= 0) { // Update the value of // ans and sum ans += sum + 1; sum = -1; } } } // Update the value of res res = Math.Min(res, ans); } // Print the value of res Console.Write(res); } // Driver Code public static void Main() { List< int > a = new List< int >(){ 1, -3, 1, 0 }; minOperations(a); } } // This code is contributed by bgangwar59. |
Javascript
<script> // JavaScript program for the above approach // Function to find minimum operations // needed to make the product of any // two adjacent elements in prefix // sum array negative function minOperations(a) { // Stores the minimum operations let res =Number.MAX_VALUE; let N = a.length; for (let r = 0; r < 2; r++) { // Stores the prefix sum // and number of operations let sum = 0, ans = 0; // Traverse the array for (let i = 0; i < N; i++) { // Update the value of sum sum += a[i]; // Check if i+r is odd if ((i + r) % 2) { // Check if prefix sum // is not positive if (sum <= 0) { // Update the value of // ans and sum ans += -sum + 1; sum = 1; } } else { // Check if prefix sum is // not negative if (sum >= 0) { // Update the value of // ans and sum ans += sum + 1; sum = -1; } } } // Update the value of res res = Math.min(res, ans); } // Print the value of res document.write(res); } // Driver Code let a = [1, -3, 1, 0 ]; minOperations(a); // This code is contributed by Potta Lokesh </script> |
4
Time Complexity: O(N)
Auxiliary Space: O(1)