659. Split Array into Consecutive Subsequences

You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Return whether you can make such a split.

Example 1:

Input: [1,2,3,3,4,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3
3, 4, 5

Example 2:

Input: [1,2,3,3,4,4,5,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3, 4, 5
3, 4, 5

Example 3:

Input: [1,2,3,4,4,5]
Output: False

Note:

1. The length of the input is in range of [1, 10000]

Approach #1: C++.

class Solution {
public:
    bool isPossible(vector<int>& nums) {
        int pre = nums[0] - 1;
        int a1 = 0, a2 = 0, a3 = 0;
        for (int i = 0; i < nums.size(); ) {
            int j = i;
            while (j+1 < nums.size() && nums[j+1] == nums[j]) ++j;
            int cnt = j - i + 1;
            int cur = nums[i];
            if (cur != pre + 1) {
                if (a1 != 0 || a2 != 0) return false;
                a3 = 0;
                a1 += cnt;
            } else {
                int b1 = 0, b2 = 0, b3 = 0;
                if (a1 > cnt) return false;
                b2 += a1, cnt -= a1, a1 = 0;
                if (a2 > cnt) return false;
                b3 += a2, cnt -= a2, a2 = 0;
                b3 += min(a3, cnt), cnt -= min(cnt, a3);
                a1 = cnt;
                a2 = b2;
                a3 = b3;
            }
            
            pre = cur;
            i = j + 1;
            
        }
        
        return a1 == 0 && a2 == 0;
    }
};

　　

Analysis:

In this problem we use a1, a2, a3 represent the number of the subsequences with the length of 1, 2, 3.

cnt represent the number of same elements in this loop.

pre represent the number in the last time loop we force on (nums[i-1]).

first : we should judge if the array is consequent with the pre number and cur number. if so, we continue the next step, otherwise, we should judge if a1 and a2 equal to 0.

second : we should put the cur number in to the previous subsequences with the length of 1 or 2. if at this loop the same numbers (cnt) smaller than a1 or a2, this means that in the next loop we will have subsequences' length less than 3, so we should return false; otherwise, we update the value of a1, a2 and a3.

finlly : we judge if a1 == 0 and a2 == 0.

Approach #2: Java.

class Solution {
    public boolean isPossible(int[] nums) {
        int pre = Integer.MIN_VALUE, p1 = 0, p2 = 0, p3 = 0;
        int cur = 0, cnt = 0, c1 = 0, c2 = 0, c3 = 0;
        
        for (int i = 0; i < nums.length; pre = cur, p1 = c1, p2 = c2, p3 = c3) {
            for (cur = nums[i], cnt = 0; i < nums.length && cur == nums[i]; cnt++, i++);
            if (cur != pre + 1) {
                if (p1 != 0 || p2 != 0) return false;
                c1 = cnt; c2 = 0; c3 = 0;
            } else {
                if (cnt < p1 + p2) return false;
                c1 = Math.max(0, cnt - (p1 + p2 + p3));
                c2 = p1;
                c3 = p2 + Math.min(p3, cnt - (p1 + p2));
            }
        }
        
        return (p1 == 0 && p2 == 0);
    }
}