Given an array **arr[]** of size **N**, the task is to find the minimum number of increment or decrement operations required at any index i such that for each **i** **(1 ≤ i < N)** if the sum of elements at index from **1 **to **i** is **positive **then the sum of elements from **1 **to **i + 1** must be **negative **or vice versa.

**Note:** Consider array as 1-based indexing.

**Examples:**

Input:arr[] = {3, -4, 5, 0, 1}Output:6Explanation:

Convert the array as {3, -4, 5, -5, 2}. Here, the sum of elements till i is represented as s_{i.}

For i = 1, s_{1}= 3 and s_{2}= 3 + (-4) = -1. s_{1}is positive and s_{2}is negative.

For i=2, s_{2}= -1 and s_{3}= 3 + (-4) + 5 = 4. s_{2}is negative and s_{3}is positive.

For i = 3, s_{3}= 4 and s_{4}= 3 + (-4) + 5 + (-5) = -1. s_{3}is positive and s_{4}is negative.

For i = 4, s_{4}= -1 and s_{5}= 3 + (-4) + 5 +(-5) + 2 = 1. s_{4}is negative and s_{5}is positive.

Input:arr[] = {1, -2, 2, -3}Output:0Explanation:

Given array already satisfies the condition. Therefore, no need to perform any operation.

**Approach:** The array will satisfy the conditions if for each i from **1 to N – 1**:

- If i is
**odd**, then the sum of elements from 1 to i is**positive**. - If i is
**even**, then**negative**and vice versa.

Try both the above possibilities and choose the one which gives the minimum number of operations. Below are the steps:

- Initialize a variable
*num_of_ops* - For any index
**i**, if**i**is**even**and the sum of elements from 1 to i**negative**, then add**(1+|sum|)**in the**arr[i]**to make it**positive**. Now the sum of elements from 1 to i will be**1**. Also add**(1+|sum|)**in the*num_of_ops i.e., to count the number of operations.* - If
**i**is**odd**and the sum of elements from 1 to i is**positive**, then subtract**(1+|sum|)**from**a[i]**to make it**negative**. Now the sum of elements from 1 to i will be**-1.**Also add**(1+|sum|)**in the*num_of_ops. i.e., to count the number of operations.* - Similarly, find the number of operations taking for even i, the sum of elements till i is negative and for odd i sum of elements till i is positive.
- Choose the minimum number of operations from the above two procedures.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <iostream>` `using` `namespace` `std;` `// Function to find minimum number` `// of operations to get desired array` `int` `minOperations(` `int` `a[], ` `int` `N)` `{` ` ` `int` `num_of_ops1, num_of_ops2, sum;` ` ` `num_of_ops1 = num_of_ops2 = sum = 0;` ` ` `// For even 'i', sum of` ` ` `// elements till 'i' is negative` ` ` `// For odd 'i', sum of` ` ` `// elements till 'i' is positive` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `sum += a[i];` ` ` `// If i is even and sum is positive,` ` ` `// make it negative by subtracting` ` ` `// 1 + |s| from a[i]` ` ` `if` `(i % 2 == 0 && sum >= 0) {` ` ` `num_of_ops1 += (1 + ` `abs` `(sum));` ` ` `sum = -1;` ` ` `}` ` ` `// If i is odd and sum is negative,` ` ` `// make it positive by` ` ` `// adding 1 + |s| into a[i]` ` ` `else` `if` `(i % 2 == 1 && sum <= 0) {` ` ` `num_of_ops1 += (1 + ` `abs` `(sum));` ` ` `sum = 1;` ` ` `}` ` ` `}` ` ` `sum = 0;` ` ` `// For even 'i', the sum of` ` ` `// elements till 'i' is positive` ` ` `// For odd 'i', sum of` ` ` `// elements till 'i' is negative` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `sum += a[i];` ` ` `// Check if 'i' is odd and sum is` ` ` `// positive, make it negative by` ` ` `// subtracting 1 + |s| from a[i]` ` ` `if` `(i % 2 == 1 && sum >= 0) {` ` ` `num_of_ops2 += (1 + ` `abs` `(sum));` ` ` `sum = -1;` ` ` `}` ` ` `// Check if 'i' is even and sum` ` ` `// is negative, make it positive` ` ` `// by adding 1 + |s| into a[i]` ` ` `else` `if` `(i % 2 == 0 && sum <= 0) {` ` ` `num_of_ops2 += (1 + ` `abs` `(sum));` ` ` `sum = 1;` ` ` `}` ` ` `}` ` ` `// Return the minimum of the two` ` ` `return` `min(num_of_ops1, num_of_ops2);` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given array arr[]` ` ` `int` `arr[] = { 3, -4, 5, 0, 1 };` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Function Call` ` ` `cout << minOperations(arr, N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to find minimum number` `// of operations to get desired array` `static` `int` `minOperations(` `int` `a[], ` `int` `N)` `{` ` ` `int` `num_of_ops1, num_of_ops2, sum;` ` ` `num_of_ops1 = num_of_ops2 = sum = ` `0` `;` ` ` `// For even 'i', sum of` ` ` `// elements till 'i' is negative` ` ` `// For odd 'i', sum of` ` ` `// elements till 'i' is positive` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++)` ` ` `{` ` ` `sum += a[i];` ` ` `// If i is even and sum is positive,` ` ` `// make it negative by subtracting` ` ` `// 1 + |s| from a[i]` ` ` `if` `(i % ` `2` `== ` `0` `&& sum >= ` `0` `)` ` ` `{` ` ` `num_of_ops1 += (` `1` `+ Math.abs(sum));` ` ` `sum = -` `1` `;` ` ` `}` ` ` `// If i is odd and sum is negative,` ` ` `// make it positive by` ` ` `// adding 1 + |s| into a[i]` ` ` `else` `if` `(i % ` `2` `== ` `1` `&& sum <= ` `0` `)` ` ` `{` ` ` `num_of_ops1 += (` `1` `+ Math.abs(sum));` ` ` `sum = ` `1` `;` ` ` `}` ` ` `}` ` ` `sum = ` `0` `;` ` ` `// For even 'i', the sum of` ` ` `// elements till 'i' is positive` ` ` `// For odd 'i', sum of` ` ` `// elements till 'i' is negative` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++)` ` ` `{` ` ` `sum += a[i];` ` ` `// Check if 'i' is odd and sum is` ` ` `// positive, make it negative by` ` ` `// subtracting 1 + |s| from a[i]` ` ` `if` `(i % ` `2` `== ` `1` `&& sum >= ` `0` `)` ` ` `{` ` ` `num_of_ops2 += (` `1` `+ Math.abs(sum));` ` ` `sum = -` `1` `;` ` ` `}` ` ` `// Check if 'i' is even and sum` ` ` `// is negative, make it positive` ` ` `// by adding 1 + |s| into a[i]` ` ` `else` `if` `(i % ` `2` `== ` `0` `&& sum <= ` `0` `)` ` ` `{` ` ` `num_of_ops2 += (` `1` `+ Math.abs(sum));` ` ` `sum = ` `1` `;` ` ` `}` ` ` `}` ` ` `// Return the minimum of the two` ` ` `return` `Math.min(num_of_ops1, num_of_ops2);` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `// Given array arr[]` ` ` `int` `arr[] = { ` `3` `, -` `4` `, ` `5` `, ` `0` `, ` `1` `};` ` ` `int` `N = arr.length;` ` ` `// Function Call` ` ` `System.out.print(minOperations(arr, N));` `}` `}` `// This code is contributed by Amit Katiyar` |

## Python3

`# Python3 program for the above approach` `# Function to find minimum number` `# of operations to get desired array` `def` `minOperations(a, N):` ` ` `num_of_ops1 ` `=` `num_of_ops2 ` `=` `sum` `=` `0` `;` ` ` `# For even 'i', sum of` ` ` `# elements till 'i' is negative` ` ` `# For odd 'i', sum of` ` ` `# elements till 'i' is positive` ` ` `for` `i ` `in` `range` `(N):` ` ` `sum` `+` `=` `a[i]` ` ` `# If i is even and sum is positive,` ` ` `# make it negative by subtracting` ` ` `# 1 + |s| from a[i]` ` ` `if` `(i ` `%` `2` `=` `=` `0` `and` `sum` `>` `=` `0` `):` ` ` `num_of_ops1 ` `+` `=` `(` `1` `+` `abs` `(` `sum` `))` ` ` `sum` `=` `-` `1` ` ` `# If i is odd and sum is negative,` ` ` `# make it positive by` ` ` `# adding 1 + |s| into a[i]` ` ` `elif` `(i ` `%` `2` `=` `=` `1` `and` `sum` `<` `=` `0` `):` ` ` `num_of_ops1 ` `+` `=` `(` `1` `+` `abs` `(` `sum` `))` ` ` `sum` `=` `1` ` ` `sum` `=` `0` ` ` `# For even 'i', the sum of` ` ` `# elements till 'i' is positive` ` ` `# For odd 'i', sum of` ` ` `# elements till 'i' is negative` ` ` `for` `i ` `in` `range` `(N):` ` ` `sum` `+` `=` `a[i]` ` ` `# Check if 'i' is odd and sum is` ` ` `# positive, make it negative by` ` ` `# subtracting 1 + |s| from a[i]` ` ` `if` `(i ` `%` `2` `=` `=` `1` `and` `sum` `>` `=` `0` `):` ` ` `num_of_ops2 ` `+` `=` `(` `1` `+` `abs` `(` `sum` `))` ` ` `sum` `=` `-` `1` ` ` `# Check if 'i' is even and sum` ` ` `# is negative, make it positive` ` ` `# by adding 1 + |s| into a[i]` ` ` `elif` `(i ` `%` `2` `=` `=` `0` `and` `sum` `<` `=` `0` `):` ` ` `num_of_ops2 ` `+` `=` `(` `1` `+` `abs` `(` `sum` `))` ` ` `sum` `=` `1` ` ` `# Return the minimum of the two` ` ` `return` `min` `(num_of_ops1, num_of_ops2)` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` ` ` `# Given array arr[]` ` ` `arr ` `=` `[ ` `3` `, ` `-` `4` `, ` `5` `, ` `0` `, ` `1` `]` ` ` `N ` `=` `len` `(arr)` ` ` `# Function call` ` ` `print` `(minOperations(arr, N))` `# This code is contributed by chitranayal` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` `// Function to find minimum number` `// of operations to get desired array` `static` `int` `minOperations(` `int` `[]a, ` `int` `N)` `{` ` ` `int` `num_of_ops1, num_of_ops2, sum;` ` ` `num_of_ops1 = num_of_ops2 = sum = 0;` ` ` `// For even 'i', sum of` ` ` `// elements till 'i' is negative` ` ` `// For odd 'i', sum of` ` ` `// elements till 'i' is positive` ` ` `for` `(` `int` `i = 0; i < N; i++)` ` ` `{` ` ` `sum += a[i];` ` ` `// If i is even and sum is positive,` ` ` `// make it negative by subtracting` ` ` `// 1 + |s| from a[i]` ` ` `if` `(i % 2 == 0 && sum >= 0)` ` ` `{` ` ` `num_of_ops1 += (1 + Math.Abs(sum));` ` ` `sum = -1;` ` ` `}` ` ` `// If i is odd and sum is negative,` ` ` `// make it positive by` ` ` `// adding 1 + |s| into a[i]` ` ` `else` `if` `(i % 2 == 1 && sum <= 0)` ` ` `{` ` ` `num_of_ops1 += (1 + Math.Abs(sum));` ` ` `sum = 1;` ` ` `}` ` ` `}` ` ` `sum = 0;` ` ` `// For even 'i', the sum of` ` ` `// elements till 'i' is positive` ` ` `// For odd 'i', sum of` ` ` `// elements till 'i' is negative` ` ` `for` `(` `int` `i = 0; i < N; i++)` ` ` `{` ` ` `sum += a[i];` ` ` `// Check if 'i' is odd and sum is` ` ` `// positive, make it negative by` ` ` `// subtracting 1 + |s| from a[i]` ` ` `if` `(i % 2 == 1 && sum >= 0)` ` ` `{` ` ` `num_of_ops2 += (1 + Math.Abs(sum));` ` ` `sum = -1;` ` ` `}` ` ` `// Check if 'i' is even and sum` ` ` `// is negative, make it positive` ` ` `// by adding 1 + |s| into a[i]` ` ` `else` `if` `(i % 2 == 0 && sum <= 0)` ` ` `{` ` ` `num_of_ops2 += (1 + Math.Abs(sum));` ` ` `sum = 1;` ` ` `}` ` ` `}` ` ` `// Return the minimum of the two` ` ` `return` `Math.Min(num_of_ops1, num_of_ops2);` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` ` ` `// Given array []arr` ` ` `int` `[]arr = { 3, -4, 5, 0, 1 };` ` ` `int` `N = arr.Length;` ` ` `// Function call` ` ` `Console.Write(minOperations(arr, N));` `}` `}` `// This code is contributed by PrinciRaj1992` |

## Javascript

`<script>` `// Javascript program for the above approach` `// Function to find minimum number` `// of operations to get desired array` `function` `minOperations(a, N)` `{` ` ` `var` `num_of_ops1, num_of_ops2, sum;` ` ` `num_of_ops1 = num_of_ops2 = sum = 0;` ` ` `// For even 'i', sum of` ` ` `// elements till 'i' is negative` ` ` `// For odd 'i', sum of` ` ` `// elements till 'i' is positive` ` ` `for` `(i = 0; i < N; i++)` ` ` `{` ` ` `sum += a[i];` ` ` `// If i is even and sum is positive,` ` ` `// make it negative by subtracting` ` ` `// 1 + |s| from a[i]` ` ` `if` `(i % 2 == 0 && sum >= 0)` ` ` `{` ` ` `num_of_ops1 += (1 + Math.abs(sum));` ` ` `sum = -1;` ` ` `}` ` ` `// If i is odd and sum is negative,` ` ` `// make it positive by` ` ` `// adding 1 + |s| into a[i]` ` ` `else` `if` `(i % 2 == 1 && sum <= 0)` ` ` `{` ` ` `num_of_ops1 += (1 + Math.abs(sum));` ` ` `sum = 1;` ` ` `}` ` ` `}` ` ` `sum = 0;` ` ` `// For even 'i', the sum of` ` ` `// elements till 'i' is positive` ` ` `// For odd 'i', sum of` ` ` `// elements till 'i' is negative` ` ` `for` `(i = 0; i < N; i++)` ` ` `{` ` ` `sum += a[i];` ` ` `// Check if 'i' is odd and sum is` ` ` `// positive, make it negative by` ` ` `// subtracting 1 + |s| from a[i]` ` ` `if` `(i % 2 == 1 && sum >= 0)` ` ` `{` ` ` `num_of_ops2 += (1 + Math.abs(sum));` ` ` `sum = -1;` ` ` `}` ` ` `// Check if 'i' is even and sum` ` ` `// is negative, make it positive` ` ` `// by adding 1 + |s| into a[i]` ` ` `else` `if` `(i % 2 == 0 && sum <= 0)` ` ` `{` ` ` `num_of_ops2 += (1 + Math.abs(sum));` ` ` `sum = 1;` ` ` `}` ` ` `}` ` ` `// Return the minimum of the two` ` ` `return` `Math.min(num_of_ops1, num_of_ops2);` `}` `// Driver Code` `// Given array arr` `var` `arr = [ 3, -4, 5, 0, 1 ];` `var` `N = arr.length;` `// Function Call` `document.write(minOperations(arr, N));` `// This code is contributed by aashish1995` `</script>` |

**Output:**

6

**Time Complexity:** O(N)**Auxiliary Space:** O(1)

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