• Array
  • 217. Contains Duplicate
  • 242. Valid Anagram
  • 1. Two Sum
  • 49. Group Anagrams
  • 347. Top K Frequent Elements
  • 238. Product of Array Except Self
  • 36. Valid Sudoku
  • 128. Longest Consecutive Sequence
• Two Pointers
  • 125.Valid Palindrome
  • 167.Two Sum II - Input Array Is Sorted
  • 15.3Sum
  • 11.Container With Most Water
  • 42.Trapping Rain Water

Easy

This problem is about finding if any number shows up more than once in a list. We use a hashmap to keep track of each number. If we see a number that's already in the hashmap, it means there's a duplicate.

The time complexity is O(n), which means we go through the list just once. The space complexity is also O(n) because we need space to store each unique number in the hashmap.

function containsDuplicate(nums) {
    let map = new Map();
    
    for (let i = 0; i < nums.length; i++) {
        if (map.has(nums[i])) {
            return true;
        }
        map.set(nums[i], true);
    }

    return false;
};

• Key Concept: Check for duplicates in a list.
• Approach: Use a hashmap to track each number's presence.
• Key Phrases:
  • "Finding if a number appears more than once"
  • "Using a hashmap for tracking"
  • "Time complexity is O(n), space complexity is O(n)"
• Example Code Snippet:

if (map.has(nums[i])) return true;
map.set(nums[i], true);

Easy

To check if two strings are anagrams(meaning they are made of the same characters in a different order), we can use two hashmaps to record the occurrence count of each character in both strings. After that, we compare the two hashmaps to see if they are identical.

The time complexity of O(n) and the space complexity of O(n). Iterates through each character of both strings once and uses two hashmaps to store their character count.

function isAnagram(s, t) {
    if (s.length !== t.length) {
        return false;
    }

    const countA = {};
    const countB = {};

    for (let i = 0; i < s.length; i++) {
        countA[s[i]] = 1 + (countA[s[i]] || 0);
        countB[t[i]] = 1 + (countB[t[i]] || 0);
    }

    const entriesA = Object.entries(countA);
    // const entriesB = Object.entries(countB);

    for (const [key, value] of entriesA) {
        if (countB[key] !== value) {
            return false;
        }
    }

    return true;
};

• Key Concept: Check if two strings have the same characters in different orders.
• Approach: Use two hashmaps to track the count of each character in both strings, then compare these counts.
• Key Phrases:
  • "Determining if strings are anagrams with hashmaps"
  • "Counting character occurrences in each string"
  • "Equal character counts indicate anagrams"
• Example Code Snippet:

countA[s[i]] = (countA[s[i]] || 0) + 1;
if (countB[key] !== value) return false;

Easy

We use a hashmap to store each number along with its index, where the number is the key and the index is the value. We iterate through the array to find the difference between the target and the current number. If this difference is already in the hashmap, it means we have found a pair that adds up to the target, and we return their indices.

The time complexity is O(n) because we only need to go through the array once. The space complexity is also O(n) as we create a hashmap to store each number and its index.

var twoSum = function(nums, target) {
    let obj = {};

    for (let i = 0; i < nums.length; i++) {
        const diff = target - nums[i];
        
        if (obj[diff] !== undefined) {
            return [i, obj[diff]];
        }

        obj[nums[i]] = i;
    }

    return [];
    
};

• Key Concept: Find two numbers that add up to a specific target.
• Approach: Store each number and its index in a hashmap, checking for the complement of each number.
• Key Phrases:
  • "Using a hashmap to store numbers and indices"
  • "Finding two numbers that sum up to the target"
  • "Iterate once, time and space complexity O(n)"
• Example Code Snippet:

const diff = target - nums[i];
if (obj[diff] !== undefined) return [i, obj[diff]];

Medium

To group anagrams, we use a hashmap where the key is a unique identifier generated from the frequency of each letter in a string. We achieve this by using a fixed-size array of 26 elements to represent the frequency of each letter (assuming lowercase English letters). Each position in the array corresponds to a letter, and we increment the count as we encounter each letter in the strings.

/**
 * @param {string[]} strs
 * @return {string[][]}
 */
var groupAnagrams = function(strs) {
    let obj = {};
    
    for (let i = 0; i < strs.length; i++) {
        const count = new Array(26).fill(0);
        for (let s of strs[i]) {
            count[s.charCodeAt(0) - 'a'.charCodeAt(0)] += 1;
        }
        const key = count.join('#');

        if (!obj[key]) {   
            obj[key] = [];
        }
        obj[key].push(strs[i])

    }

    return Object.values(obj);
};

• Key Concept: Grouping strings that are anagrams into the same category.
• Approach: Use a hashmap with keys generated from each string's character frequency.
• Key Phrases:
  • "Grouping anagrams using a hashmap."
  • "Unique key from letter frequencies."
  • "Fixed-size array for character count."
  • "each position in the array corresponds to a letter"
  • "each position be a number representing the frequency of the letter"
  • "Time complexity is O(n), space complexity is O(n)."
• Example Code Snippet:

const count = new Array(26).fill(0); // Frequency array for each letter
count[s.charCodeAt(0) - 'a'.charCodeAt(0)]++; // Increment letter count
const key = count.join('#'); // Generate unique key for the hashmap
obj[key] = (obj[key] || []).concat(strs[i]); // Group anagrams together

Medium

To identify the top k frequent elements, we first utilize a hashmap, with each element as a key and its frequency as the value. Next, we convert this information into an array, where each index represents a frequency, and we place the elements into positions that match their frequencies. Finally, by iterating through the array from the highest frequencies to the lowest, we extract the top k frequent numbers.

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number[]}
 */
var topKFrequent = function(nums, k) {
    const frequentMap = new Map();
    const counts = new Array(nums.length + 1).fill().map(() => [])
    
    for (let i = 0; i < nums.length; i++) {
        frequentMap.set(nums[i], (frequentMap.get(nums[i]) || 0) + 1);
    }

    for (let [num, frequent] of frequentMap) {
        counts[frequent].push(num);
    }    

    let res = [];

    for (let j = counts.length - 1; j >= 0; j--) {
        if (counts[j].length > 0) {
            if (k === 0) break;
            for (let c of counts[j]) {
                k -= 1;
                res.push(c);
            }
        }
    }

    return res;
};

• Key Concept: Record frequencies with a map and then find the top k by converting to an array.
• Approach: Count element frequencies using a hashmap, sort or iterate these frequencies, and then select the top k elements.
• Key Phrases:
  • "Using a hashmap to track element frequencies."
  • "Converting frequency data into an array."
  • "Array indices represent frequencies."
  • "Iterating from array find the top k number"
• Example Code Snippet:

// Record each element's frequency in a hashmap
frequentMap.set(nums[i], (frequentMap.get(nums[i]) || 0) + 1);

// Convert frequencies into an array index, storing elements at these indices
counts[frequent].push(num);

// Retrieve the top k frequent elements by traversing the array from the end
for (let c of counts[j]) {
    if (k === 0) break;
    res.push(c);
    k--;
}

Medium

To solve the problem efficiently, we recognize that the product at each position in the array can be derived from multiplying all the elements to the left of that position by all the elements to the right. This insight allows us to avoid the brute force approach, which would involve nested loops and result in O(n^2) time complexity. Instead, we calculate the products of all elements to the left and to the right of every position, then multiply these two values to get the result for each position, ensuring O(n) time complexity. It's important to note that the first element of the prefix product and the last element of the suffix product should be 1, as there are no elements outside their respective sides.

To solve the problem efficiently, we can calculate the products of all elements to the left and right of every position, then multiply them. This way, we avoid nested loops and get O(n) time complexity. Remember that the first element of the prefix product and the last element of the suffix product are always 1.

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var productExceptSelf = function(nums) {
    const n = nums.length;
    const prefix = new Array(n);
    const suffix = new Array(n);
    const result = new Array(n);

    prefix[0] = 1;
    for (let i = 1; i < n; i++) {
        prefix[i] = prefix[i - 1] * nums[i - 1];
    }

    suffix[n - 1] = 1;
    for (let j = n - 2; j >= 0; j--) {
        suffix[j] = suffix[j + 1] * nums[j + 1];
    }

    let m = 0;
    while (m < n) {
        result[m] = prefix[m] * suffix[m];
        m++;
    }
    
    return result;
};

• Key Concept: Calculate the result for each position by multiplying the left and right products.
• Approach: Use two arrays to hold the products of all elements to the left and right of each position, then multiply these for the final result.
• Key Phrases:

  • "Avoiding brute force and nested loops."

  • "Prefix and suffix product arrays."

  • "First and last elements of product arrays are 1."

  • "Linear time complexity O(n)."

• Example Code Snippet:

// Initialize prefix and suffix products
prefix[0] = 1;
suffix[n - 1] = 1;

// Compute prefix products
for (let i = 1; i < n; i++) {
    prefix[i] = prefix[i - 1] * nums[i - 1];
}

// Compute suffix products
for (let j = n - 2; j >= 0; j--) {
    suffix[j] = suffix[j + 1] * nums[j + 1];
}

// Multiply prefix and suffix products for the result
for (let m = 0; m < n; m++) {
    result[m] = prefix[m] * suffix[m];
}

Medium

We use three hash tables to keep track of all the numbers in each row, column, and 3x3 square of the Sudoku board. The key part is figuring out the index for each square, which we do with 3 * Math.floor(i / 3) + Math.floor(j / 3). This way, we can easily check for any repeated numbers in these sections."

/**
 * @param {character[][]} board
 * @return {boolean}
 */
var isValidSudoku = function(board) {
    const rows = Array.from({ length: 9 }, () => new Set());
    const cols = Array.from({ length: 9 }, () => new Set());
    const squares = Array.from({ length: 9 }, () => new Set());

    for (let i = 0; i < board.length; i++) {
        
        for (let j = 0; j < board[i].length; j++) {
            const num = board[i][j];

            if (num === '.') 
                continue;

            const m = Math.floor(j / 3);
            const n = Math.floor(i / 3);
            const k = 3 * m + n;

            if (rows[j].has(num) || cols[i].has(num) || squares[k].has(num))
                return false;

            rows[j].add(num);
            cols[i].add(num);
            squares[k].add(num);            
        }
    }
    
    return true;
};

• Key Concept: Checking for duplicates in each row, column, and 3x3 square.
• Approach: Use three hash tables to keep track of the numbers in rows, columns, and squares.
• Key Phrases:
  • "Three hash tables for rows, columns, and squares."
  • "Calculating square index with a formula."
  • "Identifying duplicates efficiently."
• Time Complexity: O(n^2), because we iterate through each cell of the 9x9 Sudoku board.
• Space Complexity: O(n), as we maintain three hash tables for the 9 rows, 9 columns, and 9 squares, and n refers to the total number of cells on the board.

let squareIndex = 3 * Math.floor(i / 3) + Math.floor(j / 3);
if (hashTables[squareIndex].has(num)) return false;

Medium

To find the longest consecutive sequence in an array, we use a Set to hold the numbers as it helps speed up the search process. We start counting for a sequence only if the current number is the beginning of a sequence, meaning there's no previous consecutive number. Searching for an element within an array inside a loop can lead to O(n^2) time complexity.

/**
 * @param {number[]} nums
 * @return {number}
 */
var longestConsecutive = function(nums) {
    let longest = 0;
    let set = new Set(nums);
    
    for (let i = 0; i < nums.length; i++) {
        if (!set.has(nums[i] - 1)) {
            
            let long = 0;
            
            while(set.has(nums[i] + long)) {
                long += 1;
            }

            longest = Math.max(long, longest);
        }
    } 

    return longest;
};

• Key Concept: Identifying if the current element is the start of a consecutive sequence.
• Approach: Use a Set to store the elements for quicker searches, starting the count for a sequence only when an element doesn't have a previous consecutive number.
• Key Phrases:
  • "Using a Set for efficient search."
  • "Counting starts from the sequence beginning."
  • "Avoiding O(n^2) complexity with array searches."

let set = new Set(nums);
for (let num of nums) {
    if (!set.has(num - 1)) {
        // Start counting from this number
    }
}

Easy

To check if a given string is a palindrome (reads the same forward and backward), this solution uses a two-pointer technique. The two pointers start at the beginning

The main idea is to validate characters as letters or numbers before comparing them. If both characters are valid and identical when converted to lowercase, the pointers continue to move inward. If at any point the characters do not match, the function returns false, indicating the string is not a palindrome.

Time Complexity: O(n), where n is the length of the string, because each character is checked at most twice.

Space Complexity: O(1), since no additional space proportional to the input size is used.

/**  
 * @param {string} s  
 * @return {boolean}  
 */
var isPalindrome = function (s) {  
    let left = 0, right = s.length - 1;  
  
    while (left < right) {  
        const leftChar = s[left].toLowerCase();  
        const rightChar = s[right].toLowerCase();  
  
        if (!isValid(leftChar)) {  
            left++;  
            continue;  
        }  
  
        if (!isValid(rightChar)) {  
            right--;  
            continue;  
        }  
  
        if (leftChar === rightChar) {  
            left++;  
            right--;  
        } else {  
            return false;  
        }  
    }  
  
    return true;  
};  
  
function isValid(char) {  
  return ('0' <= char && char <= '9') || ('a' <= char && char <= 'z') || ('A' <= char && char <= 'Z');  
}

• Key Concept: Verifying palindrome status by comparing characters from both ends of a string.
• Approach: Use two pointers to compare characters, skipping letters and numbers characters.
• Key Phrases:
  • "Two-pointer technique
  • "Skipping letters and numbers characters."
• Example Code Snippet:

if (!isValid(leftChar)) left++;
if (leftChar !== rightChar) return false;

Medium

Given a sorted array with no duplicates, we can use two pointers to search for a target sum. We start with one pointer at the beginning and another at the end of the array. If the sum is greater than the target, we decrease the right pointer to find a smaller sum. If the sum is smaller than the target, we increase the left pointer to find a bigger sum. This approach allows us to efficiently find the target sum in the array.

Time Complexity: O(n), where n is the length of the array. Each element is involved in a sum calculation at most once.

Space Complexity: O(1), as the solution uses a constant amount of extra space.

/**  
 * @param {number[]} numbers  
 * @param {number} target  
 * @return {number[]}  
 */  
var twoSum = function(numbers, target) {  
  let left = 0, right = numbers.length - 1;  
  
  while (left < right) {  
    const sum = numbers[left] + numbers[right];  
  
    if (sum === target) {  
      return [left + 1, right + 1];  
    }  
  
    if (sum > target) {  
      right--;  
    } else {  
      left++;  
    }  
  }  
  
  return [];  
};

Key Concept:

• Use two pointers to find a pair of numbers that sum to a target in a sorted array.

Approach:

• Adjust two pointers based on their sum's comparison to the target value.

Key Phrases:

• "Two-pointer search method"
• "Increment left pointer for larger sums"
• "Decrement right pointer for smaller sums"
• "Efficient sum matching"
• "Pointer adjustments based on target"
• "Avoiding duplicates with sorted input"
• "Optimal pointer movement"

Example Code Snippet:

const sum = numbers[left] + numbers[right];
if (sum === target) return [left + 1, right + 1];
if (sum > target) right--;
else left++;

Medium

To begin with, we have to sort the array in question. After that, we need to iterate through the sorted array, ensuring the current element every time. We also need to make two pointers to search for the remaining elements. It's important to note that to avoid duplicate results, we must skip the same position during iteration and move the pointer.

Time Complexity: O(n^2), where n is the number of elements in the array. Sorting the array takes O(n log n), and the two-pointer search for each element takes linear time.

Space Complexity: O(1) or O(n), depending on the sorting algorithm implementation. The space is used mainly for the output array which stores the triplets.

/**  
 * @param {number[]} nums  
 * @return {number[][]}  
 */  
var threeSum = function(nums) {  
  const numbers = nums.sort((a, b) => a - b);  
  let res = [];  
  
  for (let i = 0; i < numbers.length; i++) {  
    const current = numbers[i];  
    if (current === numbers[i - 1])  
      continue;  
  
    let left = i + 1, right = numbers.length - 1;  
  
    while (left < right) {  
      const sum = current + numbers[left] + numbers[right];  
  
      if (sum > 0){  
        right--;  
      } else if (sum < 0) {  
        left++;  
      } else {  
        res.push([current, numbers[left], numbers[right]]);  
        left++;  
  
        while (left < right && numbers[left] === numbers[left - 1]) {  
          left++;  
        }  
      }  
    }  
  }  
  
  return res;  
};

Key Concept:

• Traverse each element and use two pointers to find the remaining two for a zero sum.

Approach:

• Iterate over each element, using two pointers for subsequent values. Ensure skipping duplicate elements by comparing with the previous position during each iteration and pointer movement.

Key Phrases:

• "Sort array first for searching"
• "Two-pointer search method"
• "Ensure current element uniqueness"
• "Skip duplicates during pointer adjustment"
• "Iterate with duplicate checks"
• "two-pointer traversal"
• "Zero-sum triplet discovery"
• "Combine sorting with two-pointer technique"

if (sum === 0) {
    res.push([numbers[i], numbers[left], numbers[right]]);
    while (left < right && numbers[left] === numbers[left + 1]) left++;  // Avoid duplicates in results
}

Medium

To find the largest area, we need to multiply the distance between two pointers by the minimum height. We then compare the heights of the two pointers and move the pointer with the shorter height. This is done to find the height position and keep track of the maximum area.

Time Complexity: O(n), where n is the number of elements in the array. Each element is considered exactly once when calculating the area.

Space Complexity: O(1), as only a constant amount of extra space is used besides the input list.

/**  
 * @param {number[]} height  
 * @return {number}  
 */  
var maxArea = function(height) {  
  let left = 0; right = height.length - 1;  
  let maxArea = 0;  
  
  while (left < right) {  
    const width = right - left;  
    const minHeight = Math.min(height[left], height[right]);  
  
    const area = width * minHeight;  
  
    maxArea = Math.max(maxArea, area);  
  
    // move the shorter line  
    if (height[right] < height[left]) {  
      right--;  
    } else {  
      left++;  
    }  
  
  }  
  
  return maxArea;  
};

Key Concept:

• Calculating the maximum water container area by two lines using their minimum height and the distance between them.

Approach:

• Utilize a two-pointer strategy starting at the start and end of the array and moving inwards, always moving the pointer at the shorter height to potentially increase the area.

Key Phrases:

• "Multiplying distance by minimum height"
• "Comparing heights of two pointers"
• "Moving the pointer with the shorter height"
• "Tracking maximum area throughout"
• "Dynamic pointer adjustment for optimal area"
• "Height-driven pointer movement"
• "Utilizing width and height for area calculation"
• "Strategy for maximum water containment"
• "Efficient area maximization by shifting pointers"

Example Code Snippet:

while (left < right) {
    const width = right - left;  // Calculate the distance between the two pointers
    const minHeight = Math.min(height[left], height[right]);  // Find the minimum height

    const area = width * minHeight;  // Calculate the area
    maxArea = Math.max(maxArea, area);  // Update the maximum area found so far

    // Move the pointer with the shorter height to potentially increase the area
    if (height[right] < height[left]) {
        right--;
    } else {
        left++;
    }
}

Hard

Water can be trapped at each position in the array depending on the minimum of the maximum heights of the walls on the left and right sides. We must maintain and track the maximum heights on both sides throughout the process. Using two pointers representing these maximum heights, we adjust the pointer associated with the shorter side during each iteration, calculating the trapped water by subtracting the current height from the minimum of the two maximum heights and then adding the result to a cumulative total.

Key Concept**: To find the maximum height between the left and right sides at a position, compare the heights and subtract the minimum from the current value. Repeat this process and sum up the results. The force method requires multiple iterations, while the two pointers method only needs one iteration.

Approach: To find the maximum height between the left and right sides of the pointer, use leftMax and rightMax. During the iteration process, determine which of leftMax and rightMax is smaller and shift it. While shifting, keep track of the maximum value, and add up the result of the pointer position subtracted from the max value.

/**  
 * @param {number[]} height  
 * @return {number}  
 */  
var trap = function(height) {  
  let res = 0;  
  let left = 0, right = height.length - 1;  
  let leftMax = height[left], rightMax = height[right];  
  
  while (left < right) {  
    if (leftMax < rightMax) {  
      left++;  
      leftMax = Math.max(leftMax, height[left]);  
      res += leftMax - height[left];  
    } else {  
      right--;  
      rightMax = Math.max(rightMax, height[right]);  
      res += rightMax - height[right];  
    }  
  }  
  
  return res;  
};

Key Phrases:

• "Minimum of left and right maximum heights"
• "Track max heights on both sides"
• "Adjust pointers based on the shorter side"
• "Calculate trapped water at each step"
• "Subtract current height from minimum max height"
• "Accumulate water trapped at each position"
• "Single-pass with two-pointer technique"
• "Dynamic update of leftMax and rightMax"
• "Optimize tracking by moving the shorter wall"
• "Efficient water trapping with minimal iterations"

Brute-Force Method:

• Time Complexity: O(n^2) — "Each element is compared with every other element."
• Space Complexity: O(1) — "Only constant space is used, regardless of input size."

Two-Pointer Method:

• Time Complexity: O(n) — "Each element is visited once."
• Space Complexity: O(1) — "Uses minimal space independent of the input size."

Example Code Snippet:

while (left < right) {  
  if (leftMax < rightMax) {  
    left++;  
    leftMax = Math.max(leftMax, height[left]);  // Update the maximum height on the left
    res += leftMax - height[left];  // Calculate and add trapped water at current position
  } else {  
    right--;  
    rightMax = Math.max(rightMax, height[right]);  // Update the maximum height on the right
    res += rightMax - height[right];  // Calculate and add trapped water at current position
  }  
}