🚀Algorithm Techniques for Efficient Problem Solving
Atul Tripathi

Atul Tripathi @atultrp

About: Atul Tripathi: React JS front-end developer with 3+ years of experience. Passionate about building dynamic, user-friendly interfaces using the latest web tech.

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Dec 27, 2020

🚀Algorithm Techniques for Efficient Problem Solving

Publish Date: Mar 4 '25
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🚀 Must-Know Algorithm Techniques for Efficient Problem Solving

Mastering algorithmic techniques can significantly improve your coding efficiency. Below are some key strategies along with examples and LeetCode problems to help you practice. 💡


🔹 1. Two Pointer Technique 🏃‍♂️🏃‍♀️

Concept: Use two pointers to efficiently search through a sorted list.

Common Use Cases:

  • Searching in sorted arrays
  • Finding pairs that meet a condition

Example: Find two numbers in a sorted array that sum to a target value.

function twoSumSorted(arr, target) {
  let left = 0, right = arr.length - 1;
  while (left < right) {
    let sum = arr[left] + arr[right];
    if (sum === target) return [arr[left], arr[right]];
    sum < target ? left++ : right--;
  }
  return [];
}
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Practice: Two Sum II


🔹 2. Prefix Sum ➕

Concept: Compute cumulative sums to quickly answer range sum queries.

Common Use Cases:

  • Fast range sum calculations
  • Detecting patterns in sequences

Example: Compute prefix sums for an array.

function prefixSum(arr) {
  let prefix = [0];
  for (let i = 0; i < arr.length; i++) {
    prefix[i + 1] = prefix[i] + arr[i];
  }
  return prefix;
}
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Practice: Range Sum Query


🔹 3. Top K Elements 🔝

Concept: Use sorting or heaps to find the most important elements in a list.

Example: Find the largest k elements.

function topKElements(arr, k) {
  return arr.sort((a, b) => b - a).slice(0, k);
}
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Practice: Top K Frequent Elements


🔹 4. Sliding Window 🏠

Concept: Use a moving window to optimize range-based computations.

Example: Find the maximum sum of any k consecutive elements.

function maxSumSubarray(arr, k) {
  let sum = 0, maxSum = -Infinity;
  for (let i = 0; i < k; i++) sum += arr[i];
  for (let i = k; i < arr.length; i++) {
    sum += arr[i] - arr[i - k];
    maxSum = Math.max(maxSum, sum);
  }
  return maxSum;
}
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Practice: Maximum Subarray


🔹 5. Breadth-First Search 🌳

Concept: Explore a graph layer by layer.

Example: Traverse a graph using BFS.

function bfs(graph, start) {
  let queue = [start], visited = new Set(queue);
  while (queue.length) {
    let node = queue.shift();
    console.log(node);
    for (let neighbor of graph[node]) {
      if (!visited.has(neighbor)) {
        visited.add(neighbor);
        queue.push(neighbor);
      }
    }
  }
}
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Practice: Binary Tree Level Order Traversal


🔹 6. Depth-First Search 🕵️

Concept: Explore one path deeply before backtracking.

Example: Perform DFS on a graph.

function dfs(graph, node, visited = new Set()) {
  if (visited.has(node)) return;
  visited.add(node);
  console.log(node);
  for (let neighbor of graph[node]) {
    dfs(graph, neighbor, visited);
  }
}
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Practice: Number of Islands


🔹 7. Topological Sort 📋

Concept: Order tasks when dependencies exist.

Example: Perform topological sorting.

function topologicalSort(graph) {
  let inDegree = {}, queue = [], result = [];
  Object.keys(graph).forEach(node => inDegree[node] = 0);
  Object.values(graph).flat().forEach(node => inDegree[node]++);
  Object.keys(graph).forEach(node => inDegree[node] === 0 && queue.push(node));
  while (queue.length) {
    let node = queue.shift();
    result.push(node);
    graph[node].forEach(neighbor => {
      if (--inDegree[neighbor] === 0) queue.push(neighbor);
    });
  }
  return result;
}
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Practice: Course Schedule II


🔹 8. Divide and Conquer ✂️

Concept: Break a problem into smaller parts and solve them independently.

Example: Implement merge sort.

function mergeSort(arr) {
  if (arr.length < 2) return arr;
  let mid = Math.floor(arr.length / 2);
  let left = mergeSort(arr.slice(0, mid));
  let right = mergeSort(arr.slice(mid));
  return merge(left, right);
}
function merge(left, right) {
  let result = [];
  while (left.length && right.length) {
    result.push(left[0] < right[0] ? left.shift() : right.shift());
  }
  return [...result, ...left, ...right];
}
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Practice: Sort an Array


🚀 Keep Practicing!

These techniques will significantly improve your problem-solving skills. Keep practicing and refining your approach! 💪🎉

💬 Have questions? Drop them in the comments! 👇


Comments 2 total

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  • Camila Eleano
    Camila EleanoMar 4, 2025

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