# Check whether an array can be made strictly increasing by modifying atmost one element

Given an array arr[] of positive integers, the task is to find whether it is possible to make this array strictly increasing by modifying atmost one element.

Examples:

Input: arr[] = {2, 4, 8, 6, 9, 12}
Output: Yes
By modifying 8 to 5, array will become strictly increasing.
i.e. {2, 4, 5, 6, 9, 12}

Input: arr[] = {10, 5, 2}
Output: No

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

Approach: For every element arr[i], if it is greater than both arr[i – 1] and arr[i + 1] or it is smaller than both arr[i – 1] and arr[i + 1] then arr[i] needs to be modified.
i.e. arr[i] = (arr[i – 1] + arr[i + 1]) / 2. If after modification, arr[i] = arr[i – 1] or arr[i] = arr[i + 1] then the array cannot be made strictly increasing without affecting more than a single element. Else count all such modifications, if the count of modifications in the end is less than or equal to 1 then print “Yes” else print “No”.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std;    // Function that returns true if arr[] // can be made strictly increasing after // modifying at most one element bool check(int arr[], int n) {     // To store the number of modifications     // required to make the array     // strictly increasing     int modify = 0;        // Check whether the first element needs     // to be modify or not     if (arr[0] > arr[1]) {         arr[0] = arr[1] / 2;         modify++;     }        // Loop from 2nd element to the 2nd last element     for (int i = 1; i < n - 1; i++) {            // Check whether arr[i] needs to be modified         if ((arr[i - 1] < arr[i] && arr[i + 1] < arr[i])             || (arr[i - 1] > arr[i] && arr[i + 1] > arr[i])) {                // Modifying arr[i]             arr[i] = (arr[i - 1] + arr[i + 1]) / 2;                // Check if arr[i] is equal to any of             // arr[i-1] or arr[i+1]             if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1])                 return false;                modify++;         }     }        // Check whether the last element needs     // to be modify or not     if (arr[n - 1] < arr[n - 2])         modify++;        // If more than 1 modification is required     if (modify > 1)         return false;        return true; }    // Driver code int main() {     int arr[] = { 2, 4, 8, 6, 9, 12 };     int n = sizeof(arr) / sizeof(arr[0]);        if (check(arr, n))         cout << "Yes" << endl;     else         cout << "No" << endl;        return 0; }

## Java

 // Java implementation of the approach class GFG {        // Function that returns true if arr[]     // can be made strictly increasing after     // modifying at most one element     static boolean check(int arr[], int n)     {         // To store the number of modifications         // required to make the array         // strictly increasing         int modify = 0;            // Check whether the first element needs         // to be modify or not         if (arr[0] > arr[1]) {             arr[0] = arr[1] / 2;             modify++;         }            // Loop from 2nd element to the 2nd last element         for (int i = 1; i < n - 1; i++) {                // Check whether arr[i] needs to be modified             if ((arr[i - 1] < arr[i] && arr[i + 1] < arr[i])                 || (arr[i - 1] > arr[i] && arr[i + 1] > arr[i])) {                    // Modifying arr[i]                 arr[i] = (arr[i - 1] + arr[i + 1]) / 2;                    // Check if arr[i] is equal to any of                 // arr[i-1] or arr[i+1]                 if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1])                     return false;                    modify++;             }         }            // Check whether the last element needs         // to be modify or not         if (arr[n - 1] < arr[n - 2])             modify++;            // If more than 1 modification is required         if (modify > 1)             return false;            return true;     }        // Driver code     public static void main(String[] args)     {            int[] arr = { 2, 4, 8, 6, 9, 12 };         int n = arr.length;            if (check(arr, n))             System.out.print("Yes");         else             System.out.print("No");     } }

## C#

 // C# implementation of the approach  using System;    class GFG {         // Function that returns true if arr[]      // can be made strictly increasing after      // modifying at most one element      static bool check(int []arr, int n)      {          // To store the number of modifications          // required to make the array          // strictly increasing          int modify = 0;             // Check whether the first element needs          // to be modify or not          if (arr[0] > arr[1])          {              arr[0] = arr[1] / 2;              modify++;          }             // Loop from 2nd element to the 2nd last element          for (int i = 1; i < n - 1; i++)          {                 // Check whether arr[i] needs to be modified              if ((arr[i - 1] < arr[i] && arr[i + 1] < arr[i])                  || (arr[i - 1] > arr[i] && arr[i + 1] > arr[i]))              {                     // Modifying arr[i]                  arr[i] = (arr[i - 1] + arr[i + 1]) / 2;                     // Check if arr[i] is equal to any of                  // arr[i-1] or arr[i+1]                  if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1])                      return false;                     modify++;              }          }             // Check whether the last element needs          // to be modify or not          if (arr[n - 1] < arr[n - 2])              modify++;             // If more than 1 modification is required          if (modify > 1)              return false;             return true;      }         // Driver code      public static void Main()      {             int[] arr = { 2, 4, 8, 6, 9, 12 };          int n = arr.Length;             if (check(arr, n))              Console.WriteLine("Yes");          else             Console.WriteLine("No");      }  }     // This code is contributed by AnkitRai01

## Python 3

 # Python 3 implementation of above approach    # Function that returns true if arr[] # can be made strictly increasing after # modifying at most one element def check( arr, n):        # To store the number of modifications     # required to make the array     # strictly increasing     modify = 0        # Check whether the first element needs     # to be modify or not     if (arr[0] > arr[1]) :         arr[0] = arr[1] // 2         modify+=1               # Loop from 2nd element to the 2nd last element     for i in range ( 1, n - 1):            # Check whether arr[i] needs to be modified         if ((arr[i - 1] < arr[i] and arr[i + 1] < arr[i])             or (arr[i - 1] > arr[i] and arr[i + 1] > arr[i])):                # Modifying arr[i]             arr[i] = (arr[i - 1] + arr[i + 1]) // 2                # Check if arr[i] is equal to any of             # arr[i-1] or arr[i+1]             if (arr[i] == arr[i - 1] or arr[i] == arr[i + 1]):                 return False                modify+=1                   # Check whether the last element needs     # to be modify or not     if (arr[n - 1] < arr[n - 2]):         modify+=1        # If more than 1 modification is required     if (modify > 1):         return False        return True    # Driver code if __name__ == "__main__":     arr = [ 2, 4, 8, 6, 9, 12 ]     n = len(arr)        if (check(arr, n)):         print ( "Yes")     else:         print ("No")    # This code is contributed by ChitraNayal

Output:

Yes

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Improved By : AnkitRai01, chitranayal