# Count the values greater than X in the modified array

Given an array Arr of positive intergers and a value X. The task is to find the number of values that is greater than or equal to X.

But the twist is that the values of the array are kept changing after every operation. There are two possibilies:

• If the current value is picked then all the remaining values in the array will be decreased by 1.
• If the current value is not picked then all the remaining values in the array will be increased by 1.

Examples:

Input: arr[] = {10, 5, 5, 4, 9}, X = 10
Output: 2

Explanation:
Arr = {10, 5, 5, 4, 9}, pos = 0
10 is picked

Arr = {10, 4, 4, 3, 8}, pos = 1
4 is not picked

Arr = {10, 4, 5, 4, 9}, pos = 2
5 is not picked

Arr = {10, 4, 5, 5, 10}, pos = 3
5 is not picked

Arr = {10, 4, 5, 5, 11}, pos = 4
11 is picked

Hence two elements are picked.

Input: arr[] = {5, 4, 3, 2, 1}, X = 4
Output: 1

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

Naive Approach: The idea is to iterate through every value in an array and check whether the ith value is greater, lesser, or equal than required value X. If the ith value is less than required value then increase the value from (i+1)th to end of the array by 1 else If the ith value is greater or equal than required value X then decrease the value by 1 from (i+1)th to end of the array.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count number ` `// of values greater or equal to x ` `int` `increaseDecreaseValue(``int` `arr[],  ` `                          ``int` `x, ``int` `n) ` `{ ` `    ``int` `TotalValue = 0; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``if` `(arr[i] < x)  ` `        ``{ ` ` `  `            ``// Current value is less ` `            ``// than required value ` `            ``// then we need to increase ` `            ``// the value from i + 1 to ` `            ``// len(arr) by 1 ` `            ``for` `(``int` `j = i + 1; j < n; j++) ` `            ``{ ` `                ``arr[j] += 1; ` `            ``} ` `        ``}  ` `        ``else` `        ``{ ` ` `  `            ``// Current value is greater ` `            ``// or equal to required ` `            ``// value then we need to ` `            ``// decrease the value from ` `            ``// (i + 1) to len(arr)-1 by 1 ` `            ``TotalValue += 1; ` `            ``for` `(``int` `j = i + 1; j < n; j++) ` `            ``{ ` `                ``if` `(arr[j] == 0)  ` `                ``{ ` `                    ``continue``; ` `                ``} ` `                ``else` `                ``{ ` `                    ``arr[j] -= 1; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `TotalValue; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `x = 4; ` `    ``int` `arr[] = {5, 4, 3, 2, 1}; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `countValue = increaseDecreaseValue(arr, x, n); ` `    ``cout << countValue; ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Java

 `// Java implementation of the approach  ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to count number ` `// of values greater or equal to x ` `static` `int` `increaseDecreaseValue(``int` `[]arr, ``int` `x) ` `{ ` `    ``int` `TotalValue = ``0``; ` `    ``for` `(``int` `i = ``0``; i < arr.length; i++)  ` `    ``{ ` `        ``if` `(arr[i] < x)  ` `        ``{ ` ` `  `            ``// Current value is less ` `            ``// than required value ` `            ``// then we need to increase ` `            ``// the value from i + 1 to ` `            ``// len(arr) by 1 ` `            ``for` `(``int` `j = i + ``1``; j < arr.length; j++) ` `            ``{ ` `                ``arr[j] += ``1``; ` `            ``} ` `        ``}  ` `        ``else` `        ``{ ` ` `  `            ``// Current value is greater ` `            ``// or equal to required ` `            ``// value then we need to ` `            ``// decrease the value from ` `            ``// (i + 1) to len(arr)-1 by 1 ` `            ``TotalValue += ``1``; ` `            ``for` `(``int` `j = i + ``1``; j < arr.length; j++) ` `            ``{ ` `                ``if` `(arr[j] == ``0``)  ` `                ``{ ` `                    ``continue``; ` `                ``} ` `                ``else`  `                ``{ ` `                    ``arr[j] -= ``1``; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `TotalValue; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `x = ``4``; ` `    ``int``[] arr = {``5``, ``4``, ``3``, ``2``, ``1``}; ` `    ``int` `countValue = increaseDecreaseValue(arr, x); ` `    ``System.out.println(countValue); ` `} ` `}  ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Pyhton3 implementation ` `# of the approach ` ` `  `# Function to count number ` `# of values greater or equal to x ` `def` `increaseDecreaseValue(arr, x): ` `    ``TotalValue ``=` `0` `    ``for` `i ``in` `range``(``len``(arr)): ` `        ``if` `arr[i] < x: ` `             `  `            ``# Current value is less ` `            ``# than required value ` `            ``# then we need to increase ` `            ``# the value from i + 1 to ` `            ``# len(arr) by 1 ` `            ``for` `j ``in` `range``(i ``+` `1``, ``len``(arr)): ` `                ``arr[j] ``+``=` `1` `        ``else``: ` `             `  `            ``# Current value is greater ` `            ``# or equal to required ` `            ``# value then we need to ` `            ``# decrease the value from ` `            ``# (i + 1) to len(arr)-1 by 1 ` `            ``TotalValue ``+``=` `1` `             `  `            ``for` `j ``in` `range``(i ``+` `1``, ``len``(arr)): ` `                ``if` `arr[j] ``=``=` `0``: ` `                    ``continue` `                ``arr[j] ``-``=` `1` `    ``return` `TotalValue ` ` `  ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``x ``=` `4` `    ``arr ``=` `[``5``, ``4``, ``3``, ``2``, ``1``] ` `    ``countValue ``=``\ ` `            ``increaseDecreaseValue(arr, x) ` `    ``print``(countValue) `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` ` `  `// Function to count number ` `// of values greater or equal to x ` `static` `int` `increaseDecreaseValue(``int` `[]arr, ``int` `x) ` `{ ` `    ``int` `TotalValue = 0; ` `    ``for` `(``int` `i = 0; i < arr.Length; i++)  ` `    ``{ ` `        ``if` `(arr[i] < x)  ` `        ``{ ` ` `  `            ``// Current value is less ` `            ``// than required value ` `            ``// then we need to increase ` `            ``// the value from i + 1 to ` `            ``// len(arr) by 1 ` `            ``for` `(``int` `j = i + 1; j < arr.Length; j++) ` `            ``{ ` `                ``arr[j] += 1; ` `            ``} ` `        ``}  ` `        ``else` `        ``{ ` ` `  `            ``// Current value is greater ` `            ``// or equal to required ` `            ``// value then we need to ` `            ``// decrease the value from ` `            ``// (i + 1) to len(arr)-1 by 1 ` `            ``TotalValue += 1; ` `            ``for` `(``int` `j = i + 1; j < arr.Length; j++) ` `            ``{ ` `                ``if` `(arr[j] == 0)  ` `                ``{ ` `                    ``continue``; ` `                ``} ` `                ``else` `                ``{ ` `                    ``arr[j] -= 1; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `TotalValue; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `x = 4; ` `    ``int``[] arr = {5, 4, 3, 2, 1}; ` `    ``int` `countValue = increaseDecreaseValue(arr, x); ` `    ``Console.WriteLine(countValue); ` `} ` `}  ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```1
```

ime Complexity:

Efficient Approach:

• This problem can further be optimized to .
• Here the main idea is to check by how much this index value should change.
• This can be done by using a temporary variable, here it is currentStatus that will keep the net effect on the current index by the previous decisions.
• The effect will be added to the value of that index and that will tell us the updated original value of the array.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach  ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count number ` `// of values greater or equal to x ` `int` `increaseDecreaseValue(``int` `arr[],  ` `                          ``int` `x, ``int` `n) ` `{ ` `    ``int` `TotalValue = 0; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``if` `(arr[i] < x)  ` `        ``{ ` ` `  `            ``// Current value is less ` `            ``// than required value ` `            ``// then we need to increase ` `            ``// the value from i + 1 to ` `            ``// len(arr) by 1 ` `            ``for` `(``int` `j = i + 1; j < n; j++) ` `            ``{ ` `                ``arr[j] += 1; ` `            ``} ` `        ``}  ` `        ``else` `        ``{ ` ` `  `            ``// Current value is greater ` `            ``// or equal to required ` `            ``// value then we need to ` `            ``// decrease the value from ` `            ``// (i + 1) to len(arr)-1 by 1 ` `            ``TotalValue += 1; ` `            ``for` `(``int` `j = i + 1; j < n; j++) ` `            ``{ ` `                ``if` `(arr[j] == 0)  ` `                ``{ ` `                    ``continue``; ` `                ``} ` `                ``else` `                ``{ ` `                    ``arr[j] -= 1; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `TotalValue; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `x = 4; ` `    ``int` `arr[] = {5, 4, 3, 2, 1}; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `countValue = increaseDecreaseValue(arr, x, n); ` `    ``cout << countValue; ` ` `  `    ``return` `0; ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` `         `  `    ``// Function to count number  ` `    ``// of students got selected  ` `    ``static` `int` `increaseDecreaseValue(``int` `arr[], ``int` `x) ` `    ``{ ` `        ``int` `currentStatus = ``0``; ` `        ``int` `totalValue = ``0``; ` `         `  `        ``int` `i; ` `        ``int` `len = arr.length; ` `         `  `        ``for` `(i = ``0``; i < len ; i++ ) ` `        ``{  ` `             `  `            ``// Adding currentStatus to the  ` `            ``// value of that index to get  ` `            ``// the original value  ` `             `  `            ``// if it is less than X  ` `            ``if` `(arr[i] + currentStatus < x)  ` `                ``currentStatus += ``1``; ` `             `  `            ``else` `            ``{ ` `                ``currentStatus -= ``1``; ` `                ``totalValue += ``1``; ` `            ``} ` `        ``} ` `        ``return` `totalValue; ` `    ``} ` `     `  `    ``// Driver Code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `x = ``4``; ` `        ``int` `arr[] = {``5``, ``4``, ``3``, ``2``, ``1``};  ` `        ``int` `countValue = increaseDecreaseValue(arr, x); ` `        ``System.out.println(countValue);  ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Pyhton3 implementation ` `# of the approach ` ` `  `# Function to count number ` `# of students got selected ` `def` `increaseDecreaseValue(arr, x): ` `     `  `    ``currentStatus ``=` `0` `    ``totalValue ``=` `0` `     `  `    ``for` `i ``in` `arr: ` `         `  `        ``# Adding currentStatus to the  ` `        ``# value of that index to get ` `        ``# the original value ` `         `  `        ``# if it is less than X ` `        ``if` `i ``+` `currentStatus < x: ` `            ``currentStatus ``+``=` `1` `         `  `        ``else``: ` `            ``currentStatus ``-``=` `1` `            ``totalValue ``+``=` `1` `             `  `    ``return` `totalValue ` ` `  `# Drivers Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``x ``=` `4` `    ``arr ``=` `[``5``, ``4``, ``3``, ``2``, ``1``] ` `    ``countValue ``=` `increaseDecreaseValue(arr, x) ` `    ``print``(countValue) `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// Function to count number  ` `    ``// of students got selected  ` `    ``static` `int` `increaseDecreaseValue(``int` `[]arr,  ` `                                     ``int` `x) ` `    ``{ ` `        ``int` `currentStatus = 0; ` `        ``int` `totalValue = 0; ` `         `  `        ``int` `i; ` `        ``int` `len = arr.Length; ` `         `  `        ``for` `(i = 0; i < len ; i++ ) ` `        ``{  ` `             `  `            ``// Adding currentStatus to the  ` `            ``// value of that index to get  ` `            ``// the original value  ` `             `  `            ``// if it is less than X  ` `            ``if` `(arr[i] + currentStatus < x)  ` `                ``currentStatus += 1; ` `             `  `            ``else` `            ``{ ` `                ``currentStatus -= 1; ` `                ``totalValue += 1; ` `            ``} ` `        ``} ` `        ``return` `totalValue; ` `    ``} ` `     `  `    ``// Driver Code  ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``int` `x = 4; ` `        ``int` `[]arr = {5, 4, 3, 2, 1};  ` `        ``int` `countValue = increaseDecreaseValue(arr, x); ` `        ``Console.Write(countValue);  ` `    ``} ` `} ` ` `  `// This code is contributed by ajit. `

Output:

```1
```

Time Complexity:

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