Split Array Into Consecutive Subsequences Problem


Description

LeetCode Problem 659.

You are given an integer array nums that is sorted in non-decreasing order.

Determine if it is possible to split nums into one or more subsequences such that both of the following conditions are true:

  • Each subsequence is a consecutive increasing sequence (i.e. each integer is exactly one more than the previous integer).
  • All subsequences have a length of 3 or more.

Return true if you can split nums according to the above conditions, or false otherwise.

A subsequence of an array is a new array that is formed from the original array by deleting some (can be none) of the elements without disturbing the relative positions of the remaining elements. (i.e., [1,3,5] is a subsequence of [1,2,3,4,5] while [1,3,2] is not).

Example 1:

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Input: nums = [1,2,3,3,4,5]
Output: true
Explanation: nums can be split into the following subsequences:
[1,2,3,3,4,5] --> 1, 2, 3
[1,2,3,3,4,5] --> 3, 4, 5

Example 2:

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Input: nums = [1,2,3,3,4,4,5,5]
Output: true
Explanation: nums can be split into the following subsequences:
[1,2,3,3,4,4,5,5] --> 1, 2, 3, 4, 5
[1,2,3,3,4,4,5,5] --> 3, 4, 5

Example 3:

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Input: nums = [1,2,3,4,4,5]
Output: false
Explanation: It is impossible to split nums into consecutive increasing subsequences of length 3 or more.

Constraints:

  • 1 <= nums.length <= 10^4
  • -1000 <= nums[i] <= 1000
  • nums is sorted in non-decreasing order.


Sample C++ Code

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class Solution {
public:
    bool isPossible(vector<int>& A) {
        unordered_map<int, int> left, end;
        for (int i: A) {
            left[i]++;
        }
        for (int i: A) {
            if (left[i] == 0) continue;
            left[i]--;
            if (end[i-1] > 0) {   
                end[i-1]--;
                end[i]++;
            } else if (left[i+1] > 0 && left[i+2] > 0) {
                left[i+1]--;
                left[i+2]--;
                end[i+2]++;
            } else {   
                return false;
            }
        }
        return true;
    }
};




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