Leetcode Contains Duplicate II problem solution YASH PAL, 31 July 2024 In this Leetcode Contains Duplicate II problem solution we have given an integer array nums and an integer k, return true if there are two distinct indices i and j in the array such that nums[i] == nums[j] and abs(i – j) <= k. Problem solution in Python. class Solution: def containsDuplicate(self, nums): """ :type nums: List[int] :rtype: bool """ return len(list(set(nums))) < len(nums) Problem solution in Java. class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { int end = 0; for (int i = 0; i < nums.length - 1; i++) { end = i + 1; while (end < nums.length && nums[end] != nums[i]) { end++; } //no duplicate value for nums[i] if (end >= nums.length) continue; if (nums[end] == nums[i] && end - i <= k) { return true; } } return false; } } Problem solution in C++. class Solution { public: bool containsNearbyDuplicate(vector<int>& nums, int k) { unordered_set<int> set; for(int i=0; i<nums.size(); i++) { if(set.count(nums[i])>0) return true; set.insert(nums[i]); if(set.size()>k) { set.erase(nums[i-k]); } } return false; } }; Problem solution in C. typedef struct Node{ int latest_index; int key; int val; struct Node *next; }; #define SIZE 1000 int getPos(int n){ long long num=n; if(num<0) num=-1*num; return num%SIZE; } bool containsNearbyDuplicate(int* nums, int numsSize, int k){ struct Node** hashMap=calloc(SIZE+1, sizeof(struct Node *)); for(int i=0;i<numsSize;i++){ int pos=getPos(nums[i]); struct Node* iter=hashMap[pos]; struct Node* prev=hashMap[pos]; while(iter){ if(iter->key==nums[i]){ if(iter->val>=1){ if(i <= iter->latest_index+k){ return true; } } iter->latest_index=i; iter->val++; break; } prev=iter; iter=iter->next; } if(iter==NULL){ struct Node *tmp=malloc(sizeof(struct Node)); tmp->key=nums[i]; tmp->val=1; tmp->latest_index=i; tmp->next=NULL; if(prev){ prev->next=iter; } else{ hashMap[pos]=tmp; } } } return false; } coding problems