Algorithms for Developers — Sorting, Searching & Big O

Big O Notation

• O(1) — Constant: array access, hash map lookup
• O(log n) — Logarithmic: binary search
• O(n) — Linear: single loop
• O(n log n) — Linearithmic: merge sort, quicksort
• O(n²) — Quadratic: nested loops
• O(2ⁿ) — Exponential: brute force recursion

Binary Search

function binarySearch(arr, target) {
  let left = 0, right = arr.length - 1;
  while (left <= right) {
    const mid = Math.floor((left + right) / 2);
    if (arr[mid] === target) return mid;
    if (arr[mid] < target) left = mid + 1;
    else right = mid - 1;
  }
  return -1;
}
// O(log n) — halves the search space each step

Sorting Algorithms

// Bubble Sort — O(n²)
function bubbleSort(arr) {
  for (let i = 0; i < arr.length; i++) {
    for (let j = 0; j < arr.length - i - 1; j++) {
      if (arr[j] > arr[j + 1]) {
        [arr[j], arr[j + 1]] = [arr[j + 1], arr[j]];
      }
    }
  }
  return arr;
}

// Quick Sort — O(n log n) average
function quickSort(arr) {
  if (arr.length <= 1) return arr;
  const pivot = arr[arr.length - 1];
  const left = arr.filter(x => x < pivot);
  const right = arr.filter(x => x > pivot);
  return [...quickSort(left), pivot, ...quickSort(right)];
}

Recursion

function factorial(n) {
  if (n <= 1) return 1;
  return n * factorial(n - 1);
}

function fibonacci(n) {
  if (n <= 1) return n;
  return fibonacci(n - 1) + fibonacci(n - 2);
}

// Memoized fibonacci
function fibMemo(n, memo = {}) {
  if (n in memo) return memo[n];
  if (n <= 1) return n;
  memo[n] = fibMemo(n - 1, memo) + fibMemo(n - 2, memo);
  return memo[n];
}

Dynamic Programming

// Climbing Stairs
function climbStairs(n) {
  if (n <= 2) return n;
  const dp = [0, 1, 2];
  for (let i = 3; i <= n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2];
  }
  return dp[n];
}

// Longest Common Subsequence
function lcs(s1, s2) {
  const m = s1.length, n = s2.length;
  const dp = Array(m + 1).fill(null).map(() => Array(n + 1).fill(0));
  for (let i = 1; i <= m; i++) {
    for (let j = 1; j <= n; j++) {
      if (s1[i - 1] === s2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
      else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
    }
  }
  return dp[m][n];
}