673. Number of Longest Increasing Subsequence.md

673. Number of Longest Increasing Subsequence

Question

Given an unsorted array of integers, find the number of longest increasing subsequence.

Example 1:

Input: [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequence are [1, 3, 4, 7] and [1, 3, 5, 7].

Example 2:

Input: [2,2,2,2,2]
Output: 5
Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5.

Note: Length of the given array will be not exceed 2000 and the answer is guaranteed to be fit in 32-bit signed int.

Solution

• Method 1: dp
  • dp[i]: up to current index, the max number of increasing subset.
  • count[i]: the number of ways to keep the maximum dp number up to index i.
    • update the maximum: we update the count to the previous ones, since we only need to append current value to all previous ones.
    • if max is the same, we need to append current value to all previous ways.

  class Solution {
      public int findNumberOfLIS(int[] nums) {
          int len = nums.length;
          if(len == 0) return 0;
          int[] dp = new int[len];
          int[] count = new int[len];
          Arrays.fill(dp, 1);
          Arrays.fill(count, 1);
          int max = 1;
          for(int i = 1; i < len; i++){
              for(int j = i - 1; j >= 0; j--){
                  if(nums[i] > nums[j] && dp[j] + 1 > dp[i]){
                      dp[i] = dp[j] + 1;
                      count[i] = count[j];
                  }else if(nums[i] > nums[j] && dp[j] + 1 == dp[i]){
                      count[i] += count[j];
                  }
              }
              max = Math.max(max, dp[i]);
          }
          int res = 0;
          for(int i = 0; i < len; i++){
              if(dp[i] == max){
                  res += count[i];
              }
          }
          return res;
      }
  }

Reference

1. 673. Number of Longest Increasing Subsequence 最长递增子序列的个数