JavaScript Program Count number of Equal Pairs in a String
Last Updated :
18 Jul, 2024
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In this article, we are going to learn how can we count a number of equal pairs in a string. Counting equal pairs in a string involves finding and counting pairs of consecutive characters that are the same. This task can be useful in various applications, including pattern recognition and data analysis.
Examples:
Input: 'pqr'
Output: 3
Explanation:
3 pairs that are equal are (p, p), (q, q) and (r, r)
Input: 'HelloWorld'
Output: 18
Table of Content
Naive Approach
The straightforward method involves using two nested loops to iterate through the string, identifying all pairs, and maintaining a count of these pairs.
Example:
// JavaScript program to determine
// the count of equal character pairs
// Function to calculate the
// count of equal character pairs
function fun(str) {
// Length of the input string
let strLength = str.length;
// Variable to store the
// count of equal character pairs
let pairCount = 0;
// Nested loops to compare
// characters for equal pairs
for (let i = 0; i < strLength; i++) {
for (let j = 0; j < strLength; j++) {
// If an equal pair is found
if (str[i] == str[j]) {
pairCount++;
}
}
}
return pairCount;
}
// Driver Code
let str = "geeksforgeeks";
console.log(fun(str));
// JavaScript program to determine// the count of equal character pairs// Function to calculate the // count of equal character pairsfunction fun(str) { // Length of the input string let strLength = str.length; // Variable to store the // count of equal character pairs let pairCount = 0; // Nested loops to compare // characters for equal pairs for (let i = 0; i < strLength; i++) { for (let j = 0; j < strLength; j++) { // If an equal pair is found if (str[i] == str[j]) { pairCount++; } } } return pairCount;}// Driver Codelet str = "geeksforgeeks";console.log(fun(str));Output
31
Time Complexity: O(n2), n is the length of input string
Space Complexity: O(1)
Efficient Appraoch
- In this approach, We must efficiently determine the count of distinct pairs of characters in linear time.
- Notably, pairs like (x, y) and (y, x) are treated as distinct.
- To accomplish this, we employ a hash table to record the occurrences of each character. If a character appears twice, it corresponds to 4 pairs: (i, i), (j, j), (i, j), and (j, i).
- By utilizing a hashing mechanism, we keep track of the frequency of each character, and for each character, the count of pairs will be the square of its frequency.
- The hash table will have a length of 256 since there are 256 distinct characters.
Example: This example shows the use of the above-explined approach.
// JavaScript program to calculate the
// count of pairs
const MAX = 256
// Function to calculate the count
// of identical pairs
function fun(str) {
// Hash table to store
// character counts
let charCount = new Array(MAX).fill(0)
// Iterate through the string and tally
// the occurrences of each character
for (let i = 0; i < str.length; i++)
charCount[str.charCodeAt(i) - 97] += 1
// Variable to hold the final count of pairs
let pairCount = 0
// Iterate through the characters and check
// for occurrences
for (let i = 0; i < MAX; i++)
pairCount += charCount[i] * charCount[i]
return pairCount
}
// Driver code
let str = "abccba"
console.log(fun(str))
// JavaScript program to calculate the// count of pairsconst MAX = 256// Function to calculate the count// of identical pairsfunction fun(str) { // Hash table to store // character counts let charCount = new Array(MAX).fill(0) // Iterate through the string and tally // the occurrences of each character for (let i = 0; i < str.length; i++) charCount[str.charCodeAt(i) - 97] += 1 // Variable to hold the final count of pairs let pairCount = 0 // Iterate through the characters and check // for occurrences for (let i = 0; i < MAX; i++) pairCount += charCount[i] * charCount[i] return pairCount}// Driver codelet str = "abccba"console.log(fun(str))Output
3
Time Complexity: O(n), n is the length of input string
Space Complexity: O(1)
Using Combinatorial Counting
Another efficient approach to count the number of equal pairs in a string is by using combinatorial counting principles. This approach leverages the fact that the number of ways to choose 2 items from n items (where order does not matter) is given by the combination formula C(n, 2) = n * (n - 1) / 2.
Example:
function countEqualPairs(s) {
let freq = {};
for (let char of s) {
if (freq[char]) {
freq[char]++;
} else {
freq[char] = 1;
}
}
let totalPairs = 0;
for (let count of Object.values(freq)) {
if (count > 1) {
totalPairs += (count * (count - 1)) / 2;
}
}
return totalPairs;
}
const inputStr = 'HelloWorld';
const output = countEqualPairs(inputStr);
console.log(output);
function countEqualPairs(s) { let freq = {}; for (let char of s) { if (freq[char]) { freq[char]++; } else { freq[char] = 1; } } let totalPairs = 0; for (let count of Object.values(freq)) { if (count > 1) { totalPairs += (count * (count - 1)) / 2; } } return totalPairs;}const inputStr = 'HelloWorld';const output = countEqualPairs(inputStr);console.log(output);Output
4
