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.
class Solution {
public int findNumberOfLIS(int[] nums) {
if (nums.length == 0)
return 0;
int dp[] = new int[nums.length];
int count[] = new int[nums.length];
int max = 1;
dp[0] = 1;
count[0] = 1;
// dp[i] -> length of LIS upto this point(including this point)
for (int i = 1; i < nums.length; i++) {
dp[i] = 1;// base case
for (int j = 0; j < i; j++) {
if (nums[j] < nums[i]) // For increasing subsequence
dp[i] = Math.max(dp[i], dp[j] + 1);
}
max = Math.max(dp[i], max);
if (dp[i] == 1)
count[i] = 1;
else {
for (int j = 0; j < i; j++) {
if (nums[j] < nums[i] && dp[i] == dp[j] + 1)
count[i] += count[j];
}
}
}
int ans = 0;
for (int i = 0; i < dp.length; i++) {
if (max == dp[i]) {
ans += count[i];
}
}
return ans;
}
}