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DXC Technology coding interview questions with detailed solutions 2025-2026. Practice DXC Technology placement coding problems, DSA questions, and programming challenges asked in recent hiring rounds.
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def two_sum(nums, target): seen = {} for i, num in enumerate(nums): complement = target - num if complement in seen: return [seen[complement], i] seen[num] = i return []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[]{};}def is_valid(s): stack = [] mapping = {')': '(', '}': '{', ']': '['} for char in s: if char in mapping: top = stack.pop() if stack else '#' if mapping[char] != top: return False else: stack.append(char) return not stackpublic boolean isValid(String s) { Stack<Character> stack = new Stack<>(); 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();}def reverse_string(s): return s[::-1]public String reverse(String s) { return new StringBuilder(s).reverse().toString();}void reverse(char str[]) { int n = strlen(str); for (int i = 0; i < n/2; i++) { char temp = str[i]; str[i] = str[n-1-i]; str[n-1-i] = temp; }}def is_palindrome(s): s = ''.join(c.lower() for c in s if c.isalnum()) return s == s[::-1]public boolean isPalindrome(String s) { s = s.toLowerCase().replaceAll("[^a-z0-9]", ""); int left = 0, right = s.length() - 1; while (left < right) { if (s.charAt(left++) != s.charAt(right--)) return false; } return true;}def merge(intervals): intervals.sort(key=lambda x: x[0]) merged = [] for interval in intervals: if not merged or merged[-1][1] < interval[0]: merged.append(interval) else: merged[-1][1] = max(merged[-1][1], interval[1]) return mergeddef max_subarray(nums): max_sum = current_sum = nums[0] for num in nums[1:]: current_sum = max(num, current_sum + num) max_sum = max(max_sum, current_sum) return max_sumpublic int maxSubArray(int[] nums) { int maxSum = nums[0], currentSum = nums[0]; for (int i = 1; i < nums.length; i++) { currentSum = Math.max(nums[i], currentSum + nums[i]); maxSum = Math.max(maxSum, currentSum); } return maxSum;}def second_largest(arr): first = second = float('-inf') for num in arr: if num > first: second = first first = num elif num > second and num != first: second = num return secondpublic int secondLargest(int[] arr) { int first = Integer.MIN_VALUE; int second = Integer.MIN_VALUE; for (int num : arr) { if (num > first) { second = first; first = num; } else if (num > second && num != first) { second = num; } } return second;}from collections import OrderedDict
class LRUCache: def __init__(self, capacity): self.cache = OrderedDict() self.capacity = capacity
def get(self, key): if key not in self.cache: return -1 self.cache.move_to_end(key) return self.cache[key]
def put(self, key, value): if key in self.cache: self.cache.move_to_end(key) self.cache[key] = value if len(self.cache) > self.capacity: self.cache.popitem(last=False)from collections import deque
def level_order(root): if not root: return [] result = [] queue = deque([root]) while queue: level = [] for _ in range(len(queue)): node = queue.popleft() level.append(node.val) if node.left: queue.append(node.left) if node.right: queue.append(node.right) result.append(level) return resultdef coin_change(coins, amount): dp = [float('inf')] * (amount + 1) dp[0] = 0 for i in range(1, amount + 1): for coin in coins: if coin <= i: dp[i] = min(dp[i], dp[i - coin] + 1) return dp[amount] if dp[amount] != float('inf') else -1Interview Experience
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Last updated: February 2026