Binary Search

perfectsquare.js

/*
1. Initialize left=1 & Right=n
2. Calculate mid = Math.floor((left + right) / 2);
2. If mid*mid = n, then n is Perfect Square
3. Else if mid*mid > n then right = mid - 1
4. Else if mid*mid < n then left = mid + 1
*/


const perfectSquare = (number) => {
  let left = 1;
  let right = number;
  while (left <= right) {
    let mid = Math.floor((left + right) / 2);
    let square = mid * mid;
    if (square === number) {
      return true;
    } else if (square < number) {
      left = mid + 1;
    } else {
      right = mid - 1;
    }
  }
  return false;
};

console.log(perfectSquare(1));

Binary Search vs Sequential Search: