334. Increasing Triplet Subsequence

Approach # 1 Brute Force [ Time Limit Exceeded ]

Algorithm

For each num of array (index from 1 to n - 1 ), we check whether index i meets such conditions

1 If it exist a index j (index from 0 to i - 1 ) Satisfies nums[j] < nums[i] - do condition 2 ,else next index i+1 .

2 If it exist a index k (index from i+1 to n ) Satisfies nums[j] > nums[i] return true ,else next index i+1 .

In the end ,if i == n return fasle .

c++

class Solution {public:    bool increasingTriplet(vector<int>& nums) {        if( nums.size() < 3) {            return false ;        }        for( int i = 1 ; i < nums.size() - 1 ; i ++ ) {            bool find = false ;            for( int j = 0 ; j < i ; j ++ ) {                if( nums[j] < nums[i] ) {                    find = true ;                    break;                }            }            if( !find ){                continue ;            }            for( int k = i + 1 ; k < nums.size() ; k ++ ) {                if( nums[k] > nums[i] ) {                    return true ;                }            }        }        return false ;    }};

Complexity Analysis

• Time complexity : O(N2).

• Space complexity : O(1).

Approach #2 Dynamic Programming [Accepted]

Algorithm

We can treat this problem as Longest Increasing Subsequence ,if the length of LIC >= 3 return true .

c++

class Solution {public:   int lengthOfLIS(vector<int>& nums) {        vector<int> LIS;        for (int i = 0; i < nums.size(); i++) {            if (LIS.size() == 0 || LIS[LIS.size() - 1] < nums[i]) {                LIS.push_back(nums[i]);            }            else {                int l = 0, r = LIS.size() - 1;                while (l < r) {                    int m = (l + r) / 2;                    if (LIS[m] >= nums[i]) {                        r = m;                    }                    else {                        l = m + 1;                    }                }                LIS[l] = nums[i];            }        }        return LIS.size();    }    bool increasingTriplet(vector<int>& nums) {        return lengthOfLIS(nums) >=3;    }};

Complexity Analysis

• Time complexity : O(N∗logN). Binary Search + Dynamic Programming

• Space complexity : O(N).

Approach #3 Scan and update [ Accepted ]

Algorithm

We scan the array from left to right and save the val used to record the first 2 elements of increasing size in an array.

c++

class Solution {public:    bool increasingTriplet(vector<int>& nums) {        if( nums.size() < 3 ) {            return false ;        }        int a = nums[0] ;        int b = INT_MAX ;        for( int i = 0 ; i < nums.size() ; i ++ ) {            if( nums[i] <= a ) {                a = nums[i] ;            }            else {                if( nums[i] <= b ) {                    b = nums[i] ;                }                else {                    return true ;                }            }        }        return false ;    }};

Complexity Analysis

• Time complexity : O(N).

• Space complexity : O(1).