Amazon Placement Papers 2024
Access 2024 Amazon placement paper questions with solutions and exam pattern analysis.
Practice Amazon technical questions with detailed solutions. Access DSA problems, system design questions, and coding interview questions from Amazon online assessment and technical interviews.
Practice with Amazon technical questions covering DSA problems, system design concepts, and coding fundamentals. These questions are representative of what you’ll encounter in Amazon’s online assessment and technical interviews.
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Amazon Placement Papers 2024
Access 2024 Amazon placement paper questions with solutions and exam pattern analysis.
Amazon Placement Papers 2025
Complete Amazon Guide
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Example: nums = [2, 7, 11, 15], target = 9. Output: [0, 1]
Code: (Java/C++)
class Solution { 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); } return new int[]{}; }}Example: nums = [1,3,-1,-3,5,3,6,7], k = 3. Output: [3,3,5,5,6,7]
Code: (Java/C++)
class Solution { public int[] maxSlidingWindow(int[] nums, int k) { int n = nums.length; if (n == 0 || k == 0) return new int[0];
int[] result = new int[n - k + 1]; Deque<Integer> deque = new ArrayDeque<>();
for (int i = 0; i < n; i++) { // Remove indices that are out of the current window if (!deque.isEmpty() && deque.peekFirst() == i - k) { deque.pollFirst(); }
// Remove indices whose corresponding values are less than nums[i] // because they will not be the maximum in the current window while (!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) { deque.pollLast(); }
deque.offerLast(i);
// The first element in the deque is the maximum for the current window if (i >= k - 1) { result[i - k + 1] = nums[deque.peekFirst()]; } } return result; }}If the character count is 1, append the character. If the character count is greater than 1, append the character followed by the count.
Example: chars = [“a”,“a”,“b”,“b”,“c”,“c”,“c”]. Output: [“a”,“2”,“b”,“2”,“c”,“3”]
Code: (Java/C++)
class Solution { public int compress(char[] chars) { int writeIndex = 0; int count = 1;
for (int i = 1; i <= chars.length; i++) { if (i < chars.length && chars[i] == chars[i - 1]) { count++; } else { chars[writeIndex++] = chars[i - 1]; if (count > 1) { String countStr = String.valueOf(count); for (char c : countStr.toCharArray()) { chars[writeIndex++] = c; } } count = 1; } } return writeIndex; }}Ready to practice more? Focus on solving these Amazon technical questions daily. Start with arrays and strings, then gradually move to graphs, trees, and system design questions. Practice coding problems in Java, C++, or Python to prepare for technical interviews.
Last updated: November 2025