Pair of similar elements at different indices in JavaScript

We are required to write a JavaScript function that takes in an array of integers as the first and the only argument.

The function is required to count the number of all such element pairs from the array that are equal in magnitude but are present at different indices.

For example, if the input array is:

const arr = [7, 9, 5, 7, 7, 5];

Then the output should be:

4

because the desired pairs are [7, 7], [7, 7], [7, 7], [5, 5]

How It Works

The algorithm uses a frequency map to track occurrences of each element. For each element, we add its current frequency to the count before incrementing the frequency. This counts all possible pairs with previous occurrences of the same element.

Example

Following is the complete implementation:

const arr = [7, 9, 5, 7, 7, 5];

const equalPairCount = (arr = []) => {
    if (!arr?.length) {
        return 0;
    }
    
    const map = {};
    let count = 0;
    
    arr.forEach((val) => {
        if (map[val]) {
            count += map[val];
        }
        map[val] = map[val] + 1 || 1;
    });
    
    return count;
};

console.log(equalPairCount(arr));

Output

4

Step-by-Step Execution

Let's trace through the algorithm with our example array [7, 9, 5, 7, 7, 5]:

const arr = [7, 9, 5, 7, 7, 5];
const map = {};
let count = 0;

// Step by step execution
arr.forEach((val, index) => {
    console.log(`Processing element ${val} at index ${index}`);
    console.log(`Current map:`, map);
    console.log(`Current count: ${count}`);
    
    if (map[val]) {
        count += map[val];
        console.log(`Added ${map[val]} pairs, new count: ${count}`);
    }
    
    map[val] = map[val] + 1 || 1;
    console.log(`Updated frequency of ${val} to ${map[val]}`);
    console.log('---');
});

console.log(`Final count: ${count}`);

Output

Processing element 7 at index 0
Current map: {}
Current count: 0
Updated frequency of 7 to 1
---
Processing element 9 at index 1
Current map: { '7': 1 }
Current count: 0
Updated frequency of 9 to 1
---
Processing element 5 at index 2
Current map: { '7': 1, '9': 1 }
Current count: 0
Updated frequency of 5 to 1
---
Processing element 7 at index 3
Current map: { '7': 1, '9': 1, '5': 1 }
Current count: 0
Added 1 pairs, new count: 1
Updated frequency of 7 to 2
---
Processing element 7 at index 4
Current map: { '7': 2, '9': 1, '5': 1 }
Current count: 1
Added 2 pairs, new count: 3
Updated frequency of 7 to 3
---
Processing element 5 at index 5
Current map: { '7': 3, '9': 1, '5': 1 }
Current count: 3
Added 1 pairs, new count: 4
Updated frequency of 5 to 2
---
Final count: 4

Alternative Approach Using Nested Loops

Here's a more intuitive but less efficient O(n²) approach:

const arr = [7, 9, 5, 7, 7, 5];

const equalPairCountBruteForce = (arr = []) => {
    let count = 0;
    
    for (let i = 0; i 

Output

Found pair: arr[0] = 7, arr[3] = 7
Found pair: arr[0] = 7, arr[4] = 7
Found pair: arr[2] = 5, arr[5] = 5
Found pair: arr[3] = 7, arr[4] = 7
Total pairs: 4


Comparison

Conclusion

The frequency map approach efficiently counts pairs of equal elements at different indices in O(n) time. It works by tracking how many times each element has been seen and adding that count to the total pairs when encountering the element again.