Contain Duplicate III*******

Given an array of integers, find out whether there are two distinct indices i and j in the array such that the difference between nums[i] and nums[j] is at most t and the difference between i and j is at most k.

Analyse:

1. Traversal each element with the remaining elements in the array. If requirements satisfied, return true.

    Time Exceeded Limited.

class Solution {
public:
    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
        for(int i = 0; i < nums.size(); i++){
            for(int j = i + 1; j < nums.size() && j - i <= k; j++){
                if(nums[j] - nums[i] <= t && nums[j] - nums[i] >= -t) return true;
            }
        }
        return false;
    }
};

class Solution {
public:
    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
        for(int i = 0; i < nums.size(); i++){
            for(int j = i + 1; j < nums.size(); j++){
                if(nums[j] - nums[i] > t || nums[j] - nums[i] < -t) continue;
                if(j - i <= k) return true;
            }
        }
        return false;
    }
};

2. 

    Runtime: 20ms.

class Solution {
public:
    static bool cmpptr(int *a, int *b){
        return *a < *b; 
    }
    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
        const int N = nums.size();
        vector<int*> numptrs(N);
        for(int i = 0; i < N; ++i){
            numptrs[i] = &nums[i];
        }
        sort(numptrs.begin(), numptrs.end(), cmpptr);
        for(int i = 0; i < N; i++){
            for(int j = i + 1; j < N; j++){
                //nums[i] and nums[j] is at most t
                if((*numptrs[j]) > (*numptrs[i]) + t) 
                    break;
                //the difference between i and j is at most k
                if(abs(numptrs[j] - numptrs[i]) <= k) return true;
            }
        }
        return false;
    }
    
};