# Delete array element in given index range [L – R]

Given an array A[] and size of array is N. The task is to delete element of array A[] are in given range L to R both are exclusive.

Examples:

Input : N = 12
A[] = { 3, 5, 3, 4, 9, 3, 1, 6, 3, 11, 12, 3}
L = 2
R = 7
Output : 3 5 3 6 3 11 12 3
since A[2] = 3 but this is exclude
A[7] =  6 this also exclude

Input : N = 10
A[] ={ 5, 8, 11, 15, 26, 14, 19, 17, 10, 14 }
L = 4
R = 6
Output :5 8 11 15 26 19 17 10 14

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

A naive approach is to delete element in range L to R with extra space.

Below is the implementation of the above approach:

## C++

 // C++ code to delete element // in given range #include using namespace std;    // Delete L to R element vector deleteElement(int A[], int L, int R, int N) {     vector B;        for (int i = 0; i < N; i++)          if (i <= L || i >= R)             B.push_back(A[i]);                    return B; }    // main Driver int main() {     int A[] = { 3, 5, 3, 4, 9, 3, 1, 6, 3, 11, 12, 3 };     int L = 2, R = 7;     int n = sizeof(A) / sizeof(A[0]);     vector res = deleteElement(A, L, R, n);     for (auto x : res)         cout << x << " ";     return 0; }

## Java

 import java.util.Vector;    // Java code to delete element  // in given range  class GFG { // Delete L to R element         static Vector deleteElement(int A[], int L, int R, int N) {         Vector B = new Vector<>();            for (int i = 0; i < N; i++) {             if (i <= L || i >= R) {                 B.add(A[i]);             }         }            return B;     }    // main Driver      public static void main(String[] args) {         int A[] = {3, 5, 3, 4, 9, 3, 1, 6, 3, 11, 12, 3};         int L = 2, R = 7;         int n = A.length;         Vector res = deleteElement(A, L, R, n);         for (Integer x : res) {             System.out.print(x + " ");         }     } } // This code is contributed by PrinciRaj1992

## Python3

 # Python 3 code to delete element # in given range    # Delete L to R element def deleteElement(A, L, R, N):     B = []        for i in range(0, N, 1):         if (i <= L or i >= R):             B.append(A[i])             return B    # Driver Code if __name__ == '__main__':     A = [3, 5, 3, 4, 9, 3, 1,              6, 3, 11, 12, 3]      L = 2     R = 7     n = len(A)     res = deleteElement(A, L, R, n)     for i in range(len(res)):         print(res[i], end = " ")    # THis code is implemented by # Surendra_Gangwar

## C#

 // C# code to delete element  // in given range  using System;  using System.Collections.Generic;     class GFG  {             // Delete L to R element      static List deleteElement(int []A,                          int L, int R, int N)      {          List B = new List();          for (int i = 0; i < N; i++)         {              if (i <= L || i >= R)              {                  B.Add(A[i]);              }          }          return B;      }         // Driver code     public static void Main()     {          int []A = {3, 5, 3, 4, 9, 3, 1, 6,                              3, 11, 12, 3};          int L = 2, R = 7;          int n = A.Length;          List res = deleteElement(A, L, R, n);          foreach (int x in res)          {              Console.Write(x + " ");          }      }  }    // This code is contributed by Rajput-Ji

## PHP

 = \$R)             \$B[] = \$A[\$i];     }     return \$B; }    // Driver Code \$A = array(3, 5, 3, 4, 9, 3, 1,               6, 3, 11, 12, 3); \$L = 2; \$R = 7; \$n = count(\$A); \$res = deleteElement(\$A, \$L, \$R, \$n); for (\$i = 0; \$i < count(\$res); \$i++) echo "\$res[\$i] ";    // This code is implemented by // Srathore ?>

Output:

3 5 3 6 3 11 12 3

Time Complexity: O(n)
Auxiliary Space : O(n)

An efficient solution without using extra space.
Below is the implementation of the above approach:

## C++

 // C++ code to delete element // in given range #include using namespace std;    // Delete L to R elements int deleteElement(int A[], int L, int R, int N) {     int i, j = 0;     for (i = 0; i < N; i++) {         if (i <= L || i >= R) {             A[j] = A[i];             j++;         }     }        // Return size of Array     // after delete element     return j;  }    // main Driver int main() {     int A[] = { 5, 8, 11, 15, 26, 14, 19, 17, 10, 14 };     int L = 2, R = 7;     int n = sizeof(A) / sizeof(A[0]);     int res_size = deleteElement(A, L, R, n);     for (int i = 0; i < res_size; i++)         cout << A[i] << " ";     return 0; }

## Java

 // Java code to delete element // in given range class GFG  {    // Delete L to R elements static int deleteElement(int A[], int L,                           int R, int N) {     int i, j = 0;     for (i = 0; i < N; i++)      {         if (i <= L || i >= R)         {             A[j] = A[i];             j++;         }     }        // Return size of Array     // after delete element     return j;  }    // Driver Code public static void main(String args[]) {     int A[] = new int[] { 5, 8, 11, 15, 26,                          14, 19, 17, 10, 14 };     int L = 2, R = 7;     int n = A.length;     int res_size = deleteElement(A, L, R, n);     for (int i = 0; i < res_size; i++)     System.out.print(A[i] + " "); } }    // This code is contributed // by Kirti_Mangal

## Python 3

 # Python 3 program  to delete element  # in given range    # Function to delete L to R element  def deleteElement(A, L, R, N) :        j = 0     for i in range(N) :         if i <= L or i >= R :             A[j] = A[i]             j += 1        # Return size of Array      # after delete element      return j        # Driver Code if __name__ == "__main__" :        A = [5, 8, 11, 15, 26, 14, 19, 17, 10, 14]     L, R = 2,7        n = len(A)     res_size = deleteElement(A, L, R, n)        for i in range(res_size) :         print(A[i],end = " ")    # This code is contributed by ANKITRAI1

## C#

 // C# code to delete element  // in given range  using System;    class GFG  {     // Delete L to R elements  static int deleteElement(int []A, int L,                           int R, int N)  {      int i, j = 0;      for (i = 0; i < N; i++)      {          if (i <= L || i >= R)          {              A[j] = A[i];              j++;          }      }         // Return size of Array      // after delete element      return j;  }     // Driver Code  public static void Main()  {      int []A = new int[] { 5, 8, 11, 15, 26,                           14, 19, 17, 10, 14 };      int L = 2, R = 7;      int n = A.Length;      int res_size = deleteElement(A, L, R, n);      for (int i = 0; i < res_size; i++)          Console.Write(A[i] + " ");  }  }     // This code is contributed by 29AjayKumar

## PHP

 = \$R)         {             \$A[\$j] = \$A[\$i];             \$j++;         }     }        // Return size of Array     // after delete element     return \$j;  }    // Driver Code \$A = array(5, 8, 11, 15, 26,             14, 19, 17, 10, 14); \$L = 2; \$R = 7; \$n = sizeof(\$A); \$res_size = deleteElement(\$A, \$L, \$R, \$n); for (\$i = 0; \$i < \$res_size; \$i++) {     echo (\$A[\$i]);     echo (" "); }     // This code is contributed  // by Shivi_Aggarwal ?>

Output:

5 8 11 17 10 14

Time Complexity: O(n)
Auxiliary Space : O(1)

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