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 indexint 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 Codeint main(){ vector<int> nums = { -3, 2, -3, 4, 2 }; // Function Call cout << minStartValue(nums); return 0;} |
Java
// Java program for the// above approachimport java.util.*;class GFG{// Function to find minimum startValue// for positive prefix sum at each indexstatic 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 Codepublic 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 indexdef 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 Codeif __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 approachusing System; class GFG{ // Function to find minimum startValue// for positive prefix sum at each indexstatic 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 Codepublic 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 indexfunction 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> |
Output:
5
Time Complexity: O(N)
Auxiliary Space: O(1)
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