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Download latest Flipkart placement papers 2025 PDF with current year online assessment questions, solutions, and updated exam patterns.
This page contains Flipkart placement papers from 2025 with current year questions, solutions, and exam patterns.
| Section | Questions | Time | Difficulty |
|---|---|---|---|
| Coding Problems | 2-3 | 60 min | Medium-Hard |
| Debugging | 1 | 30 min | Medium |
This section contains real coding questions from Flipkart placement papers 2025 based on candidate experiences from GeeksforGeeks, LeetCode, and interview forums.
Problem: Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i]. You must write an algorithm that runs in O(n) time and without using the division operator.
Example:
Input: nums = [1,2,3,4]Output: [24,12,8,6]Solution (Java):
public int[] productExceptSelf(int[] nums) { int n = nums.length; int[] result = new int[n];
// Left pass result[0] = 1; for (int i = 1; i < n; i++) { result[i] = result[i - 1] * nums[i - 1]; }
// Right pass int right = 1; for (int i = n - 1; i >= 0; i--) { result[i] *= right; right *= nums[i]; }
return result;}Time Complexity: O(n), Space Complexity: O(1) excluding output array
Problem: Design a data structure that follows the constraints of a Least Recently Used (LRU) cache. Implement the LRUCache class.
Solution (Java):
class LRUCache { class Node { int key, value; Node prev, next; Node(int k, int v) { key = k; value = v; } }
private Map<Integer, Node> cache; private int capacity; private Node head, tail;
public LRUCache(int capacity) { this.capacity = capacity; cache = new HashMap<>(); head = new Node(0, 0); tail = new Node(0, 0); head.next = tail; tail.prev = head; }
public int get(int key) { Node node = cache.get(key); if (node == null) return -1; moveToHead(node); return node.value; }
public void put(int key, int value) { Node node = cache.get(key); if (node == null) { Node newNode = new Node(key, value); cache.put(key, newNode); addToHead(newNode); if (cache.size() > capacity) { Node tail = removeTail(); cache.remove(tail.key); } } else { node.value = value; moveToHead(node); } }
private void addToHead(Node node) { node.prev = head; node.next = head.next; head.next.prev = node; head.next = node; }
private void removeNode(Node node) { node.prev.next = node.next; node.next.prev = node.prev; }
private void moveToHead(Node node) { removeNode(node); addToHead(node); }
private Node removeTail() { Node lastNode = tail.prev; removeNode(lastNode); return lastNode; }}Time Complexity: O(1) for both operations, Space Complexity: O(capacity)
Problem: Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.
Example:
Input: s = "leetcode", wordDict = ["leet","code"]Output: trueSolution (Java):
public boolean wordBreak(String s, List<String> wordDict) { Set<String> wordSet = new HashSet<>(wordDict); boolean[] dp = new boolean[s.length() + 1]; dp[0] = true;
for (int i = 1; i <= s.length(); i++) { for (int j = 0; j < i; j++) { if (dp[j] && wordSet.contains(s.substring(j, i))) { dp[i] = true; break; } } }
return dp[s.length()];}Time Complexity: O(n²), Space Complexity: O(n)
Problem: Given a reference of a node in a connected undirected graph, return a deep copy (clone) of the graph.
Solution (Java):
public Node cloneGraph(Node node) { if (node == null) return null;
Map<Node, Node> map = new HashMap<>(); Queue<Node> queue = new LinkedList<>(); queue.offer(node); map.put(node, new Node(node.val));
while (!queue.isEmpty()) { Node current = queue.poll(); for (Node neighbor : current.neighbors) { if (!map.containsKey(neighbor)) { map.put(neighbor, new Node(neighbor.val)); queue.offer(neighbor); } map.get(current).neighbors.add(map.get(neighbor)); } }
return map.get(node);}Time Complexity: O(V + E), Space Complexity: O(V)
Problem: You are given an integer array height of length n. Find two lines that together with the x-axis form a container, such that the container contains the most water.
Example:
Input: height = [1,8,6,2,5,4,8,3,7]Output: 49Solution (Java):
public int maxArea(int[] height) { int left = 0, right = height.length - 1; int maxArea = 0;
while (left < right) { int width = right - left; int area = Math.min(height[left], height[right]) * width; maxArea = Math.max(maxArea, area);
if (height[left] < height[right]) { left++; } else { right--; } }
return maxArea;}Time Complexity: O(n), Space Complexity: O(1)
Problem: A peak element is an element that is strictly greater than its neighbors. Given a 0-indexed integer array nums, find a peak element, and return its index.
Example:
Input: nums = [1,2,3,1]Output: 2Solution (Java):
public int findPeakElement(int[] nums) { int left = 0, right = nums.length - 1;
while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] > nums[mid + 1]) { right = mid; } else { left = mid + 1; } }
return left;}Time Complexity: O(log n), Space Complexity: O(1)
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