## Count all increasing subsequences

We are given an array of digits (values lie in range from 0 to 9). The task is to count all the sub sequences possible in array such that in each subsequence every digit is greater than its previous digits in the subsequence. Examples:

Input : arr[] = {1, 2, 3, 4} Output: 15 There are total increasing subsequences {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4} Input : arr[] = {4, 3, 6, 5} Output: 8 Sub-sequences are {4}, {3}, {6}, {5}, {4,6}, {4,5}, {3,6}, {3,5} Input : arr[] = {3, 2, 4, 5, 4} Output : 14 Sub-sequences are {3}, {2}, {4}, {3,4}, {2,4}, {5}, {3,5}, {2,5}, {4,5}, {3,2,5} {3,4,5}, {4}, {3,4}, {2,4}

**Method 1 (Similar to LIS)**A Simple Solution is to useDynamic Programming Solution of Longest Increasing Subsequence (LIS) problem. LikeLIS problem, we first compute count of increasing subsequences ending at every index. Finally, we return sum of all values (In LCS problem, we return max of all values).

// We count all increasing subsequences ending at every // index i subCount(i) = Count of increasing subsequences ending at arr[i]. // Like LCS, this value can be recursively computed subCount(i) = 1 + &Sun; subCount(j) where j is index of all elements such that arr[j] < arr[i] and j < i. 1 is added as every element itself is a subsequence of size 1. // Finally we add all counts to get the result. Result = ∑ subCount(i) where i varies from 0 to n-1.

**Illustration:**

For example, arr[] = {3, 2, 4, 5, 4} // There are no smaller elements on left of arr[0] // and arr[1] subCount(0) = 1 subCount(1) = 1 // Note that arr[0] and arr[1] are smaller than arr[2] subCount(2) = 1 + subCount(0) + subCount(1) = 3 subCount(3) = 1 + subCount(0) + subCount(1) + subCount(2) = 1 + 1 + 1 + 3 = 6 subCount(3) = 1 + subCount(0) + subCount(1) = 1 + 1 + 1 = 3 Result = subCount(0) + subCount(1) + subCount(2) + subCount(3) = 1 + 1 + 3 + 6 + 3 = 14.

Time Complexity : O(n^{2}) Auxiliary Space : O(n) Refer this for implementation. **Method 2 (Efficient)**The above solution doesn’t use the fact that we have only 10 possible values in given array. We can use this fact by using an array count[] such that count[d] stores current count digits smaller than d.

For example, arr[] = {3, 2, 4, 5, 4} // We create a count array and initialize it as 0. count[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0} // Note that here value is used as index to store counts count[3] += 1 = 1 // i = 0, arr[0] = 3 count[2] += 1 = 1 // i = 1, arr[0] = 2 // Let us compute count for arr[2] which is 4 count[4] += 1 + count[3] + count[2] += 1 + 1 + 1 = 3 // Let us compute count for arr[3] which is 5 count[5] += 1 + count[3] + count[2] + count[4] += 1 + 1 + 1 + 3 = 6 // Let us compute count for arr[4] which is 4 count[4] += 1 + count[0] + count[1] += 1 + 1 + 1 += 3 = 3 + 3 = 6 Note that count[] = {0, 0, 1, 1, 6, 6, 0, 0, 0, 0} Result = count[0] + count[1] + ... + count[9] = 1 + 1 + 6 + 6 {count[2] = 1, count[3] = 1 count[4] = 6, count[5] = 6} = 14.

Below is C++ implementation of above idea.

// C++ program to count increasing subsequences // in an array of digits. #include<bits/stdc++.h> using namespace std; // Function To Count all the sub-sequences // possible in which digit is greater than // all previous digits arr[] is array of n // digits int countSub(int arr[], int n) { // count[] array is used to store all sub- // sequences possible using that digit // count[] array covers all the digit // from 0 to 9 int count[10] = {0}; // scan each digit in arr[] for (int i=0; i<n; i++) { // count all possible sub-sequences by // the digits less than arr[i] digit for (int j=arr[i]-1; j>=0; j--) count[arr[i]] += count[j]; // store sum of all sub-sequences plus // 1 in count[] array count[arr[i]]++; } // now sum up the all sequences possible in // count[] array int result = 0; for (int i=0; i<10; i++) result += count[i]; return result; } // Driver program to run the test case int main() { int arr[] = {3, 2, 4, 5, 4}; int n = sizeof(arr)/sizeof(arr[0]); cout << countSub(arr,n); return 0; }

Output:

15

Time Complexity : O(n) Note that the inner loop runs at most 10 times. Auxiliary Space : O(1) Note that count has at-most 10 elements. This article is contributed by **Shashank Mishra ( Gullu )**. If you like GeeksforGeeks and would like to contribute, you can also write an article usingcontribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.