# Minimum elements to change so that for an index i all elements on the left are -ve and all elements on the right are +ve

Given an array arr of size n, the task is to find the minimum number of elements that should be changed (element value can be changed to anything) so that there exists an index 0 ≤ i ≤ n-2 such that:

1. All the numbers in range 0 to i (inclusive) are < 0.
2. All numbers in range i+1 to n-1 (inclusive) are > 0.

Examples:

Input: arr[] = {-1, -2, -3, 3, -5, 3, 4}
Output: 1
Change -5 to 5 and the array becomes {-1, -2, -3, 3, 5, 3, 4}

Input: arr[] = {3, -5}
Output: 2
Change 3 to -3 and -5 to 5

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Fix the value of i, what changes would we need to make index i the required index? Change all the positive elements on the left of i to negative and all negative elements to the right of i to positive. Hence the number of operations required would be:
(Number of positive terms on the left of i) + (Number of negative terms on the right of i)
To find the required terms, we can pre-compute them using suffix sum.
Hence, we try each i as the required index and choose the one which needs minimum changes.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum number  ` `// of required changes ` `int` `minimumChanges(``int` `n, ``int` `a[]) ` `{ ` `    ``int` `i, sf[n + 1]; ` `    ``sf[n] = 0; ` `    ``for` `(i = n - 1; i >= 0; i--) { ` `        ``sf[i] = sf[i + 1]; ` `        ``if` `(a[i] <= 0) ` `            ``sf[i]++; ` `    ``} ` ` `  `    ``// number of elements >=0 in prefix ` `    ``int` `pos = 0; ` ` `  `    ``// Minimum elements to change ` `    ``int` `mn = n; ` `    ``for` `(i = 0; i < n - 1; i++) { ` `        ``if` `(a[i] >= 0) ` `            ``pos++; ` `        ``mn = min(mn, pos + sf[i + 1]); ` `    ``} ` `    ``return` `mn; ` `} ` ` `  `// Driver Program to test above function ` `int` `main() ` `{ ` `    ``int` `a[] = { -1, -2, -3, 3, -5, 3, 4 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``cout << minimumChanges(n, a); ` `} `

 `// Java implementation of the approach ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `// Function to return the minimum number  ` `// of required changes ` `static` `int` `minimumChanges(``int` `n, ``int` `a[]) ` `{ ` `    ``int` `i; ` `    ``int` `[]sf= ``new` `int``[n+``1``]; ` `    ``sf[n] = ``0``; ` `    ``for` `(i = n - ``1``; i >= ``0``; i--) { ` `        ``sf[i] = sf[i + ``1``]; ` `        ``if` `(a[i] <= ``0``) ` `            ``sf[i]++; ` `    ``} ` ` `  `    ``// number of elements >=0 in prefix ` `    ``int` `pos = ``0``; ` ` `  `    ``// Minimum elements to change ` `    ``int` `mn = n; ` `    ``for` `(i = ``0``; i < n - ``1``; i++) { ` `        ``if` `(a[i] >= ``0``) ` `            ``pos++; ` `        ``mn = Math.min(mn, pos + sf[i + ``1``]); ` `    ``} ` `    ``return` `mn; ` `} ` ` `  `    ``// Driver Program to test above function ` `    ``public` `static` `void` `main (String[] args) { ` `    ``int` `[]a = { -``1``, -``2``, -``3``, ``3``, -``5``, ``3``, ``4` `}; ` `    ``int` `n = a.length; ` `    ``System.out.println( minimumChanges(n, a)); ` `    ``} ` `} ` `// This code is contributed by inder_verma. `

 `# Python 3 implementation of the approach ` ` `  `# Function to return the minimum  ` `# number of required changes ` `def` `minimumChanges(n, a): ` ` `  `    ``sf ``=` `[``0``] ``*` `(n ``+` `1``) ` `    ``sf[n] ``=` `0` `    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``) : ` `        ``sf[i] ``=` `sf[i ``+` `1``] ` `        ``if` `(a[i] <``=` `0``): ` `            ``sf[i] ``+``=` `1` ` `  `    ``# number of elements >=0 in prefix ` `    ``pos ``=` `0` ` `  `    ``# Minimum elements to change ` `    ``mn ``=` `n ` `    ``for` `i ``in` `range``(n ``-` `1``) : ` `        ``if` `(a[i] >``=` `0``): ` `            ``pos ``+``=` `1` `        ``mn ``=` `min``(mn, pos ``+` `sf[i ``+` `1``]) ` `     `  `    ``return` `mn ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``a ``=` `[ ``-``1``, ``-``2``, ``-``3``, ``3``, ``-``5``, ``3``, ``4` `] ` `    ``n ``=` `len``(a) ` `    ``print``(minimumChanges(n, a)) ` ` `  `# This code is contributed  ` `# by ChitraNayal `

 `using` `System; ` ` `  `// C# implementation of the approach  ` ` `  `public` `class` `GFG ` `{ ` ` `  `// Function to return the minimum number   ` `// of required changes  ` `public` `static` `int` `minimumChanges(``int` `n, ``int``[] a) ` `{ ` `    ``int` `i; ` `    ``int``[] sf = ``new` `int``[n + 1]; ` `    ``sf[n] = 0; ` `    ``for` `(i = n - 1; i >= 0; i--) ` `    ``{ ` `        ``sf[i] = sf[i + 1]; ` `        ``if` `(a[i] <= 0) ` `        ``{ ` `            ``sf[i]++; ` `        ``} ` `    ``} ` ` `  `    ``// number of elements >=0 in prefix  ` `    ``int` `pos = 0; ` ` `  `    ``// Minimum elements to change  ` `    ``int` `mn = n; ` `    ``for` `(i = 0; i < n - 1; i++) ` `    ``{ ` `        ``if` `(a[i] >= 0) ` `        ``{ ` `            ``pos++; ` `        ``} ` `        ``mn = Math.Min(mn, pos + sf[i + 1]); ` `    ``} ` `    ``return` `mn; ` `} ` ` `  `    ``// Driver Program to test above function  ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `    ``int``[] a = ``new` `int``[] {-1, -2, -3, 3, -5, 3, 4}; ` `    ``int` `n = a.Length; ` `    ``Console.WriteLine(minimumChanges(n, a)); ` `    ``} ` `} ` ` `  `// This code is contributed by Shrikant13 `

 `= 0; ``\$i``--)  ` `    ``{ ` `        ``\$sf``[``\$i``] = ``\$sf``[``\$i` `+ 1]; ` `        ``if` `(``\$a``[``\$i``] <= 0) ` `            ``\$sf``[``\$i``]++; ` `    ``} ` ` `  `    ``// number of elements >=0 in prefix ` `    ``\$pos` `= 0; ` ` `  `    ``// Minimum elements to change ` `    ``\$mn` `= ``\$n``; ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n` `- 1; ``\$i``++)  ` `    ``{ ` `        ``if` `(``\$a``[``\$i``] >= 0) ` `            ``\$pos``++; ` `        ``\$mn` `= min(``\$mn``, ``\$pos` `+ ``\$sf``[``\$i` `+ 1]); ` `    ``} ` `    ``return` `\$mn``; ` `} ` ` `  `// Driver Code ` `\$a` `= ``array``(-1, -2, -3, 3, -5, 3, 4 ); ` `\$n` `= sizeof(``\$a``); ` `echo` `minimumChanges(``\$n``, ``\$a``); ` ` `  `// This code is contributed by ajit ` `?> `

Output:
```1
```

Time Complexity : O(n)

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