10 leetcode hot problems solved(with Optimal Approaches)

10 Leetcode Hot Problems Solved

With Optimal Approaches

Discover 10 hot LeetCode problems solved with optimal approaches. Learn efficient solutions to boost coding skills and ace technical interviews.

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Introduction

If you’re preparing for FAANG interviews or product-based companies, chances are you’ve already spent hours on LeetCode. But here’s the truth: not all problems carry the same weight. Some appear again and again in coding interviews and are considered “hot problems” because of their frequency and the concepts they cover.

In this blog, we’re going to solve 10 trending LeetCode problems with the best possible solutions. Whether you are a novice programmer or an experienced developer, the following problems will firm up your data structures and algorithms (DSA) skill set and make you stand out during interviews.

10 leetcode hot problems solved

Two Sum (LeetCode – 1)

Problem: Return a list of indices of the two numbers from the given array that add up to a target value.

Optimal Approach: Keep visited numbers in a hashmap and verify if complement is present.

Time Complexity: O(n)

Space Complexity: O(n)

				
					def twoSum(nums, target):
    seen = {}
    for i, num in enumerate(nums):
        if target - num in seen:
            return [seen[target - num], i]
        seen[num] = i
console.log( 'Code is Poetry' );

Best Time to Buy and Sell Stock (LeetCode – 2)

Problem: Maximize profit by buying a day and selling a day in the future.

Optimal Approach: Keep an eye on min price so far and compute max profit.

Time Complexity: O(n)

Space Complexity: O(1)

				
					def maxProfit(prices):
    min_price, max_profit = float('inf'), 0
    for price in prices:
        min_price = min(min_price, price)
        max_profit = max(max_profit, price - min_price)
    return max_profit
console.log( 'Code is Poetry' );

Longest Substring Without Repeating Characters (LeetCode – 3)

Problem: Find the length of the longest substring with unique characters.

Optimal Solution: Employ a sliding window using a set.

Time Complexity: O(n)

Space Complexity: O(n)

				
					def lengthOfLongestSubstring(s):
    seen, l, result = set(), 0, 0
    for r in range(len(s)):
        while s[r] in seen:
            seen.remove(s[l])
            l += 1
        seen.add(s[r])
        result = max(result, r - l + 1)
    return result
console.log( 'Code is Poetry' );

Valid Parentheses (LeetCode – 4)

Problem: Validate a string with brackets.

Optimal Solution: Employ a stack to verify matching brackets.

Time Complexity: O(n)

Space Complexity: O(n)

				
					def isValid(s):
    stack, mapping = [], {')':'(', ']':'[', '}':'{'}
    for ch in s:
        if ch in mapping:
            if not stack or stack[-1] != mapping[ch]:
                return False
            stack.pop()
        else:
            stack.append(ch)
    return not stack
console.log( 'Code is Poetry' );

Merge Intervals (LeetCode – 5)

Problem: Merge all the intervals which are overlapping.

Optimal Solution: Sort intervals and merge greedily.

Time Complexity: O(n log n)

Space Complexity: O(n)

				
					def merge(intervals):
    intervals.sort(key=lambda x: x[0])
    merged = [intervals[0]]
    for start, end in intervals[1:]:
        if start <= merged[-1][1]:
            merged[-1][1] = max(merged[-1][1], end)
        else:
            merged.append([start, end])
    return merged
console.log( 'Code is Poetry' );

Search in Rotated Sorted Array

Problem: Find the target in a rotated sorted array.

Optimal Approach: Modified binary search.

Time Complexity: O(log n)

Space Complexity: O(1)

				
					def search(nums, target):
    l, r = 0, len(nums)-1
    while l <= r:
        mid = (l + r)//2
        if nums[mid] == target:
            return mid
        if nums[l] <= nums[mid]:
            if nums[l] <= target < nums[mid]:
                r = mid - 1
            else:
                l = mid + 1
        else:
            if nums[mid] < target <= nums[r]:
                l = mid + 1
            else:
                r = mid - 1
    return -1
console.log( 'Code is Poetry' );

Container With Most Water

Problem: Find the max area of two vertical lines.

Optimal Approach: Apply two-pointer technique.

Time Complexity: O(n)

Space Complexity: O(1)

				
					def maxArea(height):
    l, r, result = 0, len(height)-1, 0
    while l < r:
        result = max(result, (r-l)*min(height[l], height[r]))
        if height[l] < height[r]:
            l += 1
        else:
            r -= 1
    return result
console.log( 'Code is Poetry' );

Maximum Subarray

Problem: Find the subarray with the maximum sum that is contiguous.

Optimal Approach: Kadane’s Algorithm.

Time Complexity: O(n)

Space Complexity: O(1)

				
					def maxSubArray(nums):
    curr_sum = max_sum = nums[0]
    for num in nums[1:]:
        curr_sum = max(num, curr_sum + num)
        max_sum = max(max_sum, curr_sum)
    return max_sum
console.log( 'Code is Poetry' );

Word Break

Problem: Find whether a given string can be segmented into words using a given dictionary.

Optimal Approach: Dynamic Programming (DP).

Time Complexity: O(n²)

Space Complexity: O(n)

				
					def wordBreak(s, wordDict):
    word_set, dp = set(wordDict), [False]*(len(s)+1)
    dp[0] = True
    for i in range(1, len(s)+1):
        for j in range(i):
            if dp[j] and s[j:i] in word_set:
                dp[i] = True
                break
    return dp[-1]
console.log( 'Code is Poetry' );

Lowest Common Ancestor of a Binary Tree

Problem: Get the lowest common ancestor of two nodes in a binary tree.

Optimal Approach: Recursive DFS.

Time Complexity: O(n)

Space Complexity: O(h), where h is tree height

				
					def lowestCommonAncestor(root, p, q):
    if not root or root == p or root == q:
        return root
    left = lowestCommonAncestor(root.left, p, q)
    right = lowestCommonAncestor(root.right, p, q)
    return left if left else right
console.log( 'Code is Poetry' );

Conclusion

These 10 popular LeetCode questions touch upon virtually every fundamental concept—arrays, hashing, sliding window, stacks, binary search, two-pointer, dynamic programming, and trees. Learning them guarantees that you’ll not just nail interviews but also develop a strong foundation in problem-solving.

At Coding with IITians, we teach students DSA, system design, and interview preparation so they don’t just solve problems—but solve them efficiently.