Geeksforgeeks interview question-Longest Increasing Subsequence

Longest Increasing Subsequence

The longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order.

For example, length of LIS for {10, 22, 9, 33, 21, 50, 41, 60, 80} is 6 and LIS is {10, 22, 33, 50, 60, 80 }.


Note:
1. The selected values do not have to be consecutive and can be separated, but the order cannot be disordered.
2. The interval does not need to be calculated. Only the length of the selected number can be calculated.

Ideas:

1 starts from the minimum two.

2 when calculating the I value, you only need to compare the [I] value with all the values before I, as long as A [I] is larger than any other value A [j, then, the dynamic planning table v [I] is filled in with the maximum values of v [j] + 1 and v [I ].

V [j] + 1 indicates that the maximum incrementing subsequence before j values is v [j], and + 1 indicates adding the I value A [I], then the ascending subsequence is increased by 1.

Dynamic Programming is a good solution:

int longestIncreaseSubsequence(int A[], int n){vector
 
   v(n, 1);int maxL = 1;for (int i = 1; i < n; i++){for (int j = 0; j < i; j++){if (A[i] > A[j]) v[i] = max(v[i], v[j]+1);maxL = max(maxL, v[i]);}}return maxL;}int main(){    int arr[] = { 10, 22, 9, 33, 21, 50, 41, 10 };    int n = sizeof(arr)/sizeof(arr[0]);    printf("Length of LIS is %d\n",  longestIncreaseSubsequence( arr, n ));    system("pause");    return 0;}
 

The answer to the test example is 4.