Find first and last positions of an element in a sorted array

Given a sorted array with possibly duplicate elements, the task is to find indexes of first and last occurrences of an element x in the given array.

Examples:

Input : arr[] = {1, 3, 5, 5, 5, 5 ,67, 123, 125}    
        x = 5
Output : First Occurrence = 2
         Last Occurrence = 5

Input : arr[] = {1, 3, 5, 5, 5, 5 ,7, 123 ,125 }    
        x = 7
Output : First Occurrence = 6
         Last Occurrence = 6

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The Naive Approach is to run a for loop and check given elements in array.

1. Run a for loop and for i = 0 to n-1
2. Take first = -1 and last = -1 
3. When we find element first time then we update first = i 
4. We always update last=i whenever we find the element.
5. We print first and last.

C/C++

// C++ program to find first and last occurrence of 
// an elements in given sorted array 
#include<bits/stdc++.h> 
using namespace std; 
  
// Function for finding first and last occurrence 
// of an elements 
void findFirstAndLast(int arr[], int n, int x) 
{ 
    int first = -1, last = -1; 
    for (int i=0; i<n; i++) 
    { 
        if (x != arr[i]) 
            continue; 
        if (first == -1) 
            first = i; 
        last = i; 
    } 
    if (first != -1) 
        cout << "First Occurrence = " << first 
             << "nLast Occurrence = " << last; 
    else
        cout << "Not Found"; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = {1, 2, 2, 2, 2, 3, 4, 7 ,8 ,8 }; 
    int n = sizeof(arr)/sizeof(int); 
    int x = 8; 
    findFirstAndLast(arr, n, x); 
    return 0; 
} 

Java

// Java program to find first and last occurrence of 
// an elements in given sorted array 
import java.io.*; 
  
class GFG  
{ 
    // Function for finding first and last occurrence 
    // of an elements 
    public static void findFirstAndLast(int arr[], int x) 
    { 
        int n = arr.length; 
        int first = -1, last = -1; 
        for (int i = 0; i < n; i++) 
        { 
            if (x != arr[i]) 
                continue; 
            if (first == -1) 
                first = i; 
            last = i; 
        } 
        if (first != -1){ 
            System.out.println( "First Occurrence = "  + first); 
            System.out.println("Last Occurrence = " + last); 
        } 
        else
            System.out.println("Not Found"); 
    } 
      
    public static void main (String[] args)  
    { 
         int arr[] = {1, 2, 2, 2, 2, 3, 4, 7 ,8 ,8 }; 
         int x = 8; 
         findFirstAndLast(arr, x); 
    } 
} 

Python3

# Python 3 program to find first and  
# last occurrence of an elements in 
# given sorted array 
  
  
# Function for finding first and last  
# occurrence of an elements 
def findFirstAndLast(arr, n, x) : 
    first = -1
    last = -1
    for i in range(0,n) : 
        if (x != arr[i]) : 
            continue
        if (first == -1) : 
            first = i 
        last = i 
      
    if (first != -1) : 
        print( "First Occurrence = " , first,  
               " \nLast Occurrence = " , last) 
    else : 
        print("Not Found") 
          
          
# Driver code 
arr = [1, 2, 2, 2, 2, 3, 4, 7 ,8 ,8 ] 
n = len(arr) 
x = 8
findFirstAndLast(arr, n, x) 
      
  
# This code is contributed by Nikita Tiwari. 

C#

// C# program to find first and last 
// occurrence of an elements in given  
// sorted array 
using System; 
  
class GFG { 
      
    // Function for finding first and  
    // last occurrence of an elements 
    static void findFirstAndLast(int []arr, 
                                     int x) 
    { 
        int n = arr.Length; 
        int first = -1, last = -1; 
          
        for (int i = 0; i < n; i++) 
        { 
            if (x != arr[i]) 
                continue; 
            if (first == -1) 
                first = i; 
            last = i; 
        } 
        if (first != -1) 
        { 
            Console.WriteLine( "First " 
                 + "Occurrence = " + first); 
            Console.Write("Last "
                  + "Occurrence = " + last); 
        } 
        else
            Console.Write("Not Found"); 
    } 
      
    // Driver code 
    public static void Main ()  
    { 
        int []arr = {1, 2, 2, 2, 2, 3,  
                               4, 7 ,8 ,8 }; 
        int x = 8; 
        findFirstAndLast(arr, x); 
    } 
} 
  
// This code is contributed by nitin mittal. 

PHP

<?php 
// PHP program to find first and last 
// occurrence of an elements in given 
// sorted array 
  
// Function for finding first and last 
// occurrence of an elements 
function findFirstAndLast( $arr, $n, $x) 
{ 
    $first = -1; $last = -1; 
    for ( $i = 0; $i < $n; $i++) 
    { 
        if ($x != $arr[$i]) 
            continue; 
        if ($first == -1) 
            $first = $i; 
        $last = $i; 
    } 
    if ($first != -1) 
        echo "First Occurrence = ", $first, 
              "\n", "\nLast Occurrence = ", 
                                     $last; 
    else
        echo "Not Found"; 
} 
  
// Driver code 
    $arr = array(1, 2, 2, 2, 2, 3, 4, 
                                 7 ,8 ,8 ); 
    $n = count($arr); 
    $x = 8; 
      
    findFirstAndLast($arr, $n, $x); 
      
// This code is contributed by anuj_67. 
?> 

Output:

First Occurrence = 8
Last Occurrence = 9

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

An Efficient solution of this problem is to use binary search.
1. For the first occurrence of a number

  a) If (high >= low)
  b) Calculate  mid = low + (high - low)/2;
  c) If ((mid == 0 || x > arr[mid-1]) && arr[mid] == x)
         return mid;
  d) Else if (x > arr[mid])
        return first(arr, (mid + 1), high, x, n);
  e) Else
        return first(arr, low, (mid -1), x, n);
  f) Otherwise return -1;

2. For the last occurrence of a number

  a) if (high >= low)
  b) calculate mid = low + (high - low)/2;
  c)if( ( mid == n-1 || x < arr[mid+1]) && arr[mid] == x )
         return mid;
  d) else if(x < arr[mid])
        return last(arr, low, (mid -1), x, n);
  e) else
       return last(arr, (mid + 1), high, x, n);      
  f) otherwise return -1;

C/C++

// C++ program to find first and last occurances of 
// a number in a given sorted array 
#include<bits/stdc++.h> 
using namespace std; 
  
/* if x is present in arr[] then returns the index of 
   FIRST occurrence of x in arr[0..n-1], otherwise 
   returns -1 */
int first(int arr[], int low, int high, int x, int n) 
{ 
    if(high >= low) 
    { 
        int mid = low + (high - low)/2; 
        if( ( mid == 0 || x > arr[mid-1]) && arr[mid] == x) 
            return mid; 
        else if(x > arr[mid]) 
            return first(arr, (mid + 1), high, x, n); 
        else
            return first(arr, low, (mid -1), x, n); 
    } 
    return -1; 
} 
  
  
/* if x is present in arr[] then returns the index of 
   LAST occurrence of x in arr[0..n-1], otherwise 
   returns -1 */
int last(int arr[], int low, int high, int x, int n) 
{ 
    if (high >= low) 
    { 
        int mid = low + (high - low)/2; 
        if (( mid == n-1 || x < arr[mid+1]) && arr[mid] == x) 
            return mid; 
        else if (x < arr[mid]) 
            return last(arr, low, (mid -1), x, n); 
        else
            return last(arr, (mid + 1), high, x, n); 
    } 
    return -1; 
} 
  
// Driver program 
int main() 
{ 
    int arr[] = {1, 2, 2, 2, 2, 3, 4, 7, 8, 8}; 
    int n = sizeof(arr)/sizeof(int); 
  
    int x = 8; 
    printf("First Occurrence = %d\t", 
                     first(arr, 0, n-1, x, n)); 
    printf("\nLast Occurrence = %d\n", 
                     last(arr, 0, n-1, x, n)); 
  
    return 0; 
} 

Java

// Java program to find first and last occurrence of 
// an elements in given sorted array 
import java.io.*; 
  
class GFG  
{ 
    /* if x is present in arr[] then returns the index of 
    FIRST occurrence of x in arr[0..n-1], otherwise 
    returns -1 */
    public static int first(int arr[], int low, int high, int x, int n) 
    { 
        if(high >= low) 
        { 
            int mid = low + (high - low)/2; 
            if( ( mid == 0 || x > arr[mid-1]) && arr[mid] == x) 
                return mid; 
             else if(x > arr[mid]) 
                return first(arr, (mid + 1), high, x, n); 
            else
                return first(arr, low, (mid -1), x, n); 
        } 
    return -1; 
    } 
   
    /* if x is present in arr[] then returns the index of 
    LAST occurrence of x in arr[0..n-1], otherwise 
    returns -1 */
    public static int last(int arr[], int low, int high, int x, int n) 
    { 
        if (high >= low) 
        { 
            int mid = low + (high - low)/2; 
            if (( mid == n-1 || x < arr[mid+1]) && arr[mid] == x) 
                 return mid; 
            else if (x < arr[mid]) 
                return last(arr, low, (mid -1), x, n); 
            else
                return last(arr, (mid + 1), high, x, n); 
        } 
    return -1; 
    } 
      
    public static void main (String[] args) 
    { 
          
        int arr[] = {1, 2, 2, 2, 2, 3, 4, 7, 8, 8}; 
        int n = arr.length; 
        int x = 8; 
        System.out.println("First Occurrence = " + first(arr, 0, n-1, x, n)); 
        System.out.println("Last Occurrence = " + last(arr, 0, n-1, x, n)); 
  
    } 
} 

Python3

# Python 3 program to find first and 
# last occurances of a number in  
# a given sorted array 
  
# if x is present in arr[] then  
# returns the index of FIRST  
# occurrence of x in arr[0..n-1], 
# otherwise returns -1  
def first(arr, low, high, x, n) : 
    if(high >= low) : 
        mid = low + (high - low) // 2
        if( ( mid == 0 or x > arr[mid - 1]) and arr[mid] == x) : 
            return mid 
        elif(x > arr[mid]) : 
            return first(arr, (mid + 1), high, x, n) 
        else : 
            return first(arr, low, (mid - 1), x, n) 
      
    return -1
  
  
# if x is present in arr[] then  
# returns the index of LAST occurrence 
# of x in arr[0..n-1], otherwise 
# returns -1  
def last(arr, low, high, x, n) : 
    if (high >= low) : 
        mid = low + (high - low) // 2
        if (( mid == n - 1 or x < arr[mid + 1]) and arr[mid] == x) : 
            return mid 
        elif (x < arr[mid]) : 
            return last(arr, low, (mid - 1), x, n) 
        else : 
            return last(arr, (mid + 1), high, x, n) 
              
    return -1
      
  
# Driver program 
arr = [1, 2, 2, 2, 2, 3, 4, 7, 8, 8] 
n = len(arr) 
  
x = 8
print("First Occurrence = ", 
      first(arr, 0, n - 1, x, n)) 
print("Last Occurrence = ", 
      last(arr, 0, n - 1, x, n)) 
  
  
# This code is contributed by Nikita Tiwari. 

C#

// C# program to find first and last occurrence  
// of an elements in given sorted array 
using System; 
  
class GFG { 
      
    /* if x is present in arr[] then 
    returns the index of FIRST  
    occurrence of x in arr[0..n-1], 
    otherwise returns -1 */
    public static int first(int []arr, int low, 
                        int high, int x, int n) 
    { 
        if(high >= low) 
        { 
            int mid = low + (high - low)/2; 
              
            if( ( mid == 0 || x > arr[mid-1]) 
                             && arr[mid] == x) 
                return mid; 
            else if(x > arr[mid]) 
                return first(arr, (mid + 1), 
                                 high, x, n); 
            else
                return first(arr, low,  
                             (mid -1), x, n); 
        } 
          
        return -1; 
    } 
  
    /* if x is present in arr[] then returns 
    the index of LAST occurrence of x in  
    arr[0..n-1], otherwise returns -1 */
    public static int last(int []arr, int low,  
                        int high, int x, int n) 
    { 
        if (high >= low) 
        { 
            int mid = low + (high - low)/2; 
              
            if (( mid == n-1 || x < arr[mid+1]) 
                              && arr[mid] == x) 
                return mid; 
            else if (x < arr[mid]) 
                return last(arr, low, (mid -1), 
                                         x, n); 
            else
                return last(arr, (mid + 1), 
                                   high, x, n); 
        } 
          
        return -1; 
    } 
      
    // Driver code 
    public static void Main () 
    { 
          
        int []arr = {1, 2, 2, 2, 2, 3, 4, 7,  
                                        8, 8}; 
        int n = arr.Length; 
        int x = 8; 
        Console.WriteLine("First Occurrence = "
                   + first(arr, 0, n-1, x, n)); 
                     
        Console.Write("Last Occurrence = " + 
                      last(arr, 0, n-1, x, n)); 
    } 
} 
  
// This code is contributed by nitin mittal. 

PHP

<?php 
// PHP program to find first 
// and last occurances of a  
// number in a given sorted array 
  
// if x is present in arr[] then 
// returns the index of FIRST  
// occurrence of x in arr[0..n-1],  
// otherwise returns -1  
function first($arr, $low, $high,  
                          $x, $n) 
{ 
    if($high >= $low) 
    { 
        $mid = floor($low + ($high - $low) / 2); 
        if(($mid == 0 or $x > $arr[$mid - 1])  
                        and $arr[$mid] == $x) 
            return $mid; 
        else if($x > $arr[$mid]) 
            return first($arr, ($mid + 1), 
                           $high, $x, $n); 
        else
            return first($arr, $low, ($mid - 1), 
                                        $x, $n); 
    } 
    return -1; 
} 
  
  
// if x is present in arr[]  
// then returns the index of 
// LAST occurrence of x in  
// arr[0..n-1], otherwise 
// returns -1  
function last($arr, $low, $high, 
                        $x, $n) 
{ 
    if ($high >= $low) 
    { 
        $mid = floor($low + ($high -  
                         $low) / 2); 
        if (( $mid == $n - 1 or $x <  
            $arr[$mid + 1]) and 
            $arr[$mid] == $x) 
                return $mid; 
        else if ($x < $arr[$mid]) 
            return last($arr, $low,  
               ($mid - 1), $x, $n); 
        else
            return last($arr, ($mid + 1),  
                          $high, $x, $n); 
    return -1; 
    } 
} 
  
    // Driver Code 
    $arr = array(1, 2, 2, 2, 2, 3, 4, 7, 8, 8); 
    $n = count($arr); 
  
    $x = 8; 
    echo "First Occurrence = ", 
          first($arr, 0, $n - 1, $x, $n),"\n"; 
    echo "Last Occurrence = ", 
          last($arr, 0, $n - 1, $x, $n); 
  
// This code is contributed by anuj_67 
?> 

Output:

First Occurrence = 8    
Last Occurrence = 9

Time Complexity : O(log n)
Auxiliary Space : O(1) if we use iterative Binary Search is used (Readers can refer Binary Search article for iterative Binary Search)

Extended Problem : Count number of occurrences in a sorted array

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C/C++

Java

Python3

C#

PHP

C/C++

Java

Python3

C#

PHP

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