Minimum value to be added to the prefix sums at each array indices to make them positive
Given an array arr[] consisting of N of integers, the task is to find the minimum positive value S that needs to be added such that the prefix sum at each index of the given array after adding S is always positive.
Examples:
Input: arr[] = {-3, 2, -3, 4, 2}
Output: 5
Explanation:
For S = 5, prefix sums at each array indices are as follows:
At index 1: (5 – 3) = 2
At index 2: (2 + 2) = 4
At index 3: (4 – 3) = 1
At index 4: (1 + 4) = 5
At index 5: (5 + 2) = 7
Since all the prefix sums obtained is greater than 0, the required answer is 5.Input: arr[] = {1, 2}
Output: 1
Naive Approach: The simplest approach is to iterate over all possible values of S starting from 1, and add S to the prefix sum at each array indices and check if it is greater than 0 or not. Print that value of S for which all the prefix sums are found to be positive.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, follow the steps below:
- Initialize a variable minValue with 0 to store the minimum prefix sum.
- Initialize a variable sum to store the prefix sum at every index.
- Traverse the array arr[] over range [0, N – 1] using the variable i.
- Update the sum and add arr[i] to the sum.
- If sum is less than minValue, set minValue as sum.
- Initialize a variable startValue to store the desired result and set startValue equal to (1 – minValue).
- After the above steps, print the value of startValue as minimum possible value of S.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find minimum startValue // for positive prefix sum at each index int minStartValue(vector< int >& nums) { // Store the minimum prefix sum int minValue = 0; // Stores prefix sum at each index int sum = 0; // Traverse over the array for ( auto n : nums) { // Update the prefix sum sum += n; // Update the minValue minValue = min(minValue, sum); } int startValue = 1 - minValue; // Return the positive start value return startValue; } // Driver Code int main() { vector< int > nums = { -3, 2, -3, 4, 2 }; // Function Call cout << minStartValue(nums); return 0; } |
Java
// Java program for the // above approach import java.util.*; class GFG{ // Function to find minimum startValue // for positive prefix sum at each index static int minStartValue( int [] nums) { // Store the minimum prefix sum int minValue = 0 ; // Stores prefix sum at each index int sum = 0 ; // Traverse over the array for ( int n : nums) { // Update the prefix sum sum += n; // Update the minValue minValue = Math.min(minValue, sum); } int startValue = 1 - minValue; // Return the positive start value return startValue; } // Driver Code public static void main(String[] args) { int [] nums = { - 3 , 2 , - 3 , 4 , 2 }; // Function Call System.out.println(minStartValue(nums)); } } // This code is contributed by susmitakundugoaldanga |
Python3
# Python3 program for the # above approach # Function to find minimum startValue # for positive prefix sum at each index def minStartValue(nums): # Store the minimum prefix sum minValue = 0 # Stores prefix sum at each index sum = 0 # Traverse over the array for i in range ( len (nums)): # Update the prefix sum sum + = nums[i] # Update the minValue minValue = min (minValue, sum ) startValue = 1 - minValue # Return the positive start value return startValue # Driver Code if __name__ = = '__main__' : nums = [ - 3 , 2 , - 3 , 4 , 2 ] # Function Call print (minStartValue(nums)) # This code is contributed by Princi Singh |
C#
// C# program for the above approach using System; class GFG{ // Function to find minimum startValue // for positive prefix sum at each index static int minStartValue( int [] nums) { // Store the minimum prefix sum int minValue = 0; // Stores prefix sum at each index int sum = 0; // Traverse over the array foreach ( int n in nums) { // Update the prefix sum sum += n; // Update the minValue minValue = Math.Min(minValue, sum); } int startValue = 1 - minValue; // Return the positive start value return startValue; } // Driver Code public static void Main() { int [] nums = { -3, 2, -3, 4, 2 }; // Function Call Console.WriteLine(minStartValue(nums)); } } // This code is contributed by code_hunt |
Javascript
<script> // Javascript program for the above approach // Function to find minimum startValue // for positive prefix sum at each index function minStartValue(nums) { // Store the minimum prefix sum let minValue = 0; // Stores prefix sum at each index let sum = 0; // Traverse over the array for (let n = 0; n < nums.length; n++) { // Update the prefix sum sum += nums[n]; // Update the minValue minValue = Math.min(minValue, sum); } let startValue = 1 - minValue; // Return the positive start value return startValue; } // Driver Code let nums = [ -3, 2, -3, 4, 2 ]; // Function Call document.write(minStartValue(nums)); // This code is contributed by souravmahato348. </script> |
5
Time Complexity: O(N)
Auxiliary Space: O(1)
Contact Us