By Expert Funda Wednesday, January 25, 2023

Leetcode Question Solution by ExpertFunda

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LeetCode Solutions by ExpertFunda

Curated solutions with clean Java code, pattern explanations, and complexity analysis — built for FAANG interview success.

500+ Problems | 3 Difficulty levels | Java Solutions

#1 Easy Hash Map Array

Two Sum

Given an array of integers nums and an integer target, return the indices of the two numbers that add up to target. Each input has exactly one solution and you may not use the same element twice.

Java

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[]{map.get(complement), i};
        }
        map.put(nums[i], i);
    }
    throw new IllegalArgumentException("No solution");
}

Time: O(n) Space: O(n)

#3 Medium Sliding Window Hash Map

Longest Substring Without Repeating Characters

Given a string s, find the length of the longest substring without repeating characters. Use a sliding window with a hash map to track the last seen index of each character.

Java

public int lengthOfLongestSubstring(String s) {
    Map<Character, Integer> map = new HashMap<>();
    int left = 0, max = 0;
    for (int right = 0; right < s.length(); right++) {
        char c = s.charAt(right);
        if (map.containsKey(c)) {
            left = Math.max(map.get(c) + 1, left);
        }
        map.put(c, right);
        max = Math.max(max, right - left + 1);
    }
    return max;
}

Time: O(n) Space: O(min(m,n))

#21 Easy Linked List Two Pointers

Merge Two Sorted Lists

Merge two sorted linked lists and return the merged list. Use a dummy head node and compare values iteratively, attaching the smaller node at each step.

Java

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode cur = dummy;
    while (l1 != null && l2 != null) {
        if (l1.val < l2.val) {
            cur.next = l1;
            l1 = l1.next;
        } else {
            cur.next = l2;
            l2 = l2.next;
        }
        cur = cur.next;
    }
    cur.next = (l1 != null) ? l1 : l2;
    return dummy.next;
}

Time: O(m+n) Space: O(1)

#53 Medium Dynamic Programming Kadane’s

Maximum Subarray

Find the contiguous subarray with the largest sum. Kadane’s algorithm keeps a running sum — reset to the current element whenever the running total drops below it.

Java

public int maxSubArray(int[] nums) {
    int cur = nums[0], max = nums[0];
    for (int i = 1; i < nums.length; i++) {
        cur = Math.max(nums[i], cur + nums[i]);
        max = Math.max(max, cur);
    }
    return max;
}

Time: O(n) Space: O(1)

#20 Easy Stack Hash Map

Valid Parentheses

Given a string of brackets (, ), {, }, [, ] — determine if it is valid. Use a stack to track unmatched open brackets and a map for closing-to-opening pairs.

Java

public boolean isValid(String s) {
    Deque<Character> stack = new ArrayDeque<>();
    for (char c : s.toCharArray()) {
        if (c == '(' || c == '{' || c == '[') {
            stack.push(c);
        } else {
            if (stack.isEmpty()) return false;
            char top = stack.pop();
            if ((c == ')' && top != '(') ||
                (c == '}' && top != '{') ||
                (c == ']' && top != '['))
                return false;
        }
    }
    return stack.isEmpty();
}

Time: O(n) Space: O(n)

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