Count of subarrays which start and end with the same element

Last Updated : 12 Jul, 2025

Given an array A of size N where the array elements contain values from 1 to N with duplicates, the task is to find the total number of subarrays that start and end with the same element.


Examples: 

Input: A[] = {1, 2, 1, 5, 2} 
Output: 7 
Explanation: 
Total 7 sub-array of the given array are {1}, {2}, {1}, {5}, {2}, {1, 2, 1} and {2, 1, 5, 2} are start and end with same element.
Input: A[] = {1, 5, 6, 1, 9, 5, 8, 10, 8, 9} 
Output: 14 
Explanation: 
Total 14 sub-array {1}, {5}, {6}, {1}, {9}, {5}, {8}, {10}, {8}, {9}, {1, 5, 6, 1}, {5, 6, 1, 9, 5}, {9, 5, 8, 10, 8, 9} and {8, 10, 8} start and end with same element. 

Naive approach: For each element in the array, if it is present at a different index as well, we will increase our result by 1. Also, all 1-size subarray are part of counted in the result. Therefore, add N to the result.


Below is the implementation of the above approach: 

C++ 
// C++ program to Count total sub-array
// which start and end with same element

#include <bits/stdc++.h>
using namespace std;

// Function to find total sub-array
// which start and end with same element
void cntArray(int A[], int N)
{
    // initialize result with 0
    int result = 0;

    for (int i = 0; i < N; i++) {

        // all size 1 sub-array
        // is part of our result
        result++;

        // element at current index
        int current_value = A[i];

        for (int j = i + 1; j < N; j++) {

            // Check if A[j] = A[i]
            // increase result by 1
            if (A[j] == current_value) {
                result++;
            }
        }
    }

    // print the result
    cout << result << endl;
}

// Driver code
int main()
{
    int A[] = { 1, 5, 6, 1, 9,
                5, 8, 10, 8, 9 };
    int N = sizeof(A) / sizeof(int);

    cntArray(A, N);

    return 0;
}
Java 
// Java program to Count total sub-array
// which start and end with same element

public class Main {

    // function to find total sub-array
    // which start and end with same element
    public static void cntArray(int A[], int N)
    {
        // initialize result with 0
        int result = 0;

        for (int i = 0; i < N; i++) {

            // all size 1 sub-array
            // is part of our result
            result++;

            // element at current index
            int current_value = A[i];

            for (int j = i + 1; j < N; j++) {

                // Check if A[j] = A[i]
                // increase result by 1
                if (A[j] == current_value) {
                    result++;
                }
            }
        }

        // print the result
        System.out.println(result);
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] A = { 1, 5, 6, 1, 9,
                    5, 8, 10, 8, 9 };
        int N = A.length;
        cntArray(A, N);
    }
}
Python3 
# Python3 program to count total sub-array 
# which start and end with same element 

# Function to find total sub-array 
# which start and end with same element 
def cntArray(A, N): 

    # Initialize result with 0 
    result = 0

    for i in range(0, N): 

        # All size 1 sub-array 
        # is part of our result 
        result = result + 1

        # Element at current index 
        current_value = A[i]

        for j in range(i + 1, N): 

            # Check if A[j] = A[i] 
            # increase result by 1 
            if (A[j] == current_value): 
                result = result + 1

    # Print the result 
    print(result)
    print("\n")

# Driver code 
A = [ 1, 5, 6, 1, 9, 5, 8, 10, 8, 9 ] 
N = len(A)

cntArray(A, N)

# This code is contributed by PratikBasu
C# 
// C# program to Count total sub-array
// which start and end with same element
using System;
class GFG{

// function to find total sub-array
// which start and end with same element
public static void cntArray(int []A, int N)
{
    // initialize result with 0
    int result = 0;

    for (int i = 0; i < N; i++) 
    {

        // all size 1 sub-array
        // is part of our result
        result++;

        // element at current index
        int current_value = A[i];

        for (int j = i + 1; j < N; j++) 
        {

            // Check if A[j] = A[i]
            // increase result by 1
            if (A[j] == current_value) 
            {
                result++;
            }
        }
    }

    // print the result
    Console.Write(result);
}

// Driver code
public static void Main()
{
    int[] A = { 1, 5, 6, 1, 9,
                5, 8, 10, 8, 9 };
    int N = A.Length;
    cntArray(A, N);
}
}

// This code is contributed by Code_Mech
JavaScript 
<script>

// Javascript program to Count total sub-array
// which start and end with same element

// Function to find total sub-array
// which start and end with same element
function cntArray(A, N)
{
    // initialize result with 0
    var result = 0;

    for (var i = 0; i < N; i++) {

        // all size 1 sub-array
        // is part of our result
        result++;

        // element at current index
        var current_value = A[i];

        for (var j = i + 1; j < N; j++) {

            // Check if A[j] = A[i]
            // increase result by 1
            if (A[j] == current_value) {
                result++;
            }
        }
    }

    // print the result
    document.write( result );
}

// Driver code
var A = [1, 5, 6, 1, 9,
            5, 8, 10, 8, 9];
var N = A.length;
cntArray(A, N);

// This code is contributed by noob2000.
</script>

Output: 
14

 

Time Complexity: O(N2), where N is the size of the array
Space complexity: O(1)


Efficient approach: We can optimize the above method by observing that the answer just depends on frequencies of numbers in the original array. 
For example in array {1, 5, 6, 1, 9, 5, 8, 10, 8, 9}, frequency of 1 is 2 and sub-array contributing to answer are {1}, {1} and {1, 5, 6, 1} respectively, i.e., a total of 3. 
Therefore, calculate the frequency of each element in the array. Then for each element, the increment the count by the result yielded by the following formula: 

((frequency of element)*(frequency of element + 1)) / 2


Below is the implementation of the above approach: 

C++ 
// C++ program to Count total sub-array
// which start and end with same element

#include <bits/stdc++.h>
using namespace std;

// function to find total sub-array
// which start and end with same element
void cntArray(int A[], int N)
{
    // initialize result with 0
    int result = 0;

    // array to count frequency of 1 to N
    int frequency[N + 1] = { 0 };

    for (int i = 0; i < N; i++) {
        // update frequency of A[i]
        frequency[A[i]]++;
    }

    for (int i = 1; i <= N; i++) {
        int frequency_of_i = frequency[i];

        // update result with sub-array
        // contributed by number i
        result += +((frequency_of_i)
                    * (frequency_of_i + 1))
                  / 2;
    }

    // print the result
    cout << result << endl;
}

// Driver code
int main()
{
    int A[] = { 1, 5, 6, 1, 9, 5,
                8, 10, 8, 9 };
    int N = sizeof(A) / sizeof(int);

    cntArray(A, N);

    return 0;
}
Java 
// Java program to Count total sub-array
// which start and end with same element

public class Main {

    // function to find total sub-array which
    // start and end with same element
    public static void cntArray(int A[], int N)
    {
        // initialize result with 0
        int result = 0;

        // array to count frequency of 1 to N
        int[] frequency = new int[N + 1];

        for (int i = 0; i < N; i++) {
            // update frequency of A[i]
            frequency[A[i]]++;
        }

        for (int i = 1; i <= N; i++) {
            int frequency_of_i = frequency[i];

            // update result with sub-array
            // contributed by number i
            result += ((frequency_of_i)
                       * (frequency_of_i + 1))
                      / 2;
        }

        // print the result
        System.out.println(result);
    }

    // Driver code
    public static void main(String[] args)
    {

        int[] A = { 1, 5, 6, 1, 9, 5,
                    8, 10, 8, 9 };
        int N = A.length;
        cntArray(A, N);
    }
}
Python3 
# Python3 program to count total sub-array 
# which start and end with same element 

# Function to find total sub-array 
# which start and end with same element 
def cntArray(A, N): 

    # Initialize result with 0 
    result = 0

    # Array to count frequency of 1 to N 
    frequency = [0] * (N + 1) 

    for i in range(0, N):
        
        # Update frequency of A[i] 
        frequency[A[i]] = frequency[A[i]] + 1

    for i in range(1, N + 1): 
        frequency_of_i = frequency[i]

        # Update result with sub-array 
        # contributed by number i 
        result = result + ((frequency_of_i) *
                           (frequency_of_i + 1)) / 2

    # Print the result 
    print(int(result))
    print("\n")

# Driver code 
A = [ 1, 5, 6, 1, 9, 5, 8, 10, 8, 9 ]
N = len(A)

cntArray(A, N) 

# This code is contributed by PratikBasu
C# 
// C# program to Count total sub-array
// which start and end with same element
using System;
class GFG{

// function to find total sub-array which
// start and end with same element
public static void cntArray(int []A, int N)
{
    // initialize result with 0
    int result = 0;

    // array to count frequency of 1 to N
    int[] frequency = new int[N + 1];

    for (int i = 0; i < N; i++) 
    {
        // update frequency of A[i]
        frequency[A[i]]++;
    }

    for (int i = 1; i <= N; i++) 
    {
        int frequency_of_i = frequency[i];

        // update result with sub-array
        // contributed by number i
        result += ((frequency_of_i) * 
                   (frequency_of_i + 1)) / 2;
    }

    // print the result
    Console.Write(result);
}

// Driver code
public static void Main()
{
    int[] A = { 1, 5, 6, 1, 9, 5,
                8, 10, 8, 9 };
    int N = A.Length;
    cntArray(A, N);
}
}

// This code is contributed by Nidhi_Biet
JavaScript 
<script>

// Javascript program to Count total sub-array
// which start and end with same element

   // function to find total sub-array which
    // start and end with same element
    function cntArray(A, N)
    {
        // initialize result with 0
        let result = 0;
  
        // array to count frequency of 1 to N
        let frequency = Array.from({length: N+1}, (_, i) => 0);
  
        for (let i = 0; i < N; i++) {
            // update frequency of A[i]
            frequency[A[i]]++;
        }
  
        for (let i = 1; i <= N; i++) {
            let frequency_of_i = frequency[i];
  
            // update result with sub-array
            // contributed by number i
            result += ((frequency_of_i)
                       * (frequency_of_i + 1))
                      / 2;
        }
  
        // print the result
        document.write(result);
    }


// Driver Code
    
    let A = [ 1, 5, 6, 1, 9, 5,
                    8, 10, 8, 9 ];
        let N = A.length;
        cntArray(A, N);
      
</script>

Output: 
14

 

Time Complexity: O(N), where N is the size of the array
Space complexity: O(N)
 

Related Topic: Subarrays, Subsequences, and Subsets in Array

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Article Tags:
DSA
subarray
frequency-counting

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