![Longest increasing subsequence
Integer LongestIncreasingSubsequence( Integer[] array ) {
Integer[] lis = new Integer[array.length];
for ( int i = 0 ; i < array.length ; i++)
lis[i] = 1;
for ( int i = 0 ; i < array.length ; i++)
for ( int j = 0; j < i ; j++)
if( array[j] < array [i] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1
return max( lis ) // O(n)
}](https://image.slidesharecdn.com/longestincreasingsubsequence-210412154326/85/Longest-increasing-subsequence-21-320.jpg)
Longest increasing subsequence
- 1. LONGEST INCREASING SUBSEQUENCE Algorithm Design Course TA Sessions By S. Daneshvar
- 2. Longest increasing subsequence Find the length of the longest subsequence of a given array such that all elements of the array are sorted in an increasing order. For instance: Input: {10, 22, 9, 33, 21, 50, 41, 60} Subsequences: {10}, {22}, … , {10, 9, 33} ,{33, 21 , 50, 60}, etc. Increasing Subsequence: {10}, {22}, … , {33, 50, 60} , … ,{ 10, 22, 33, 50 , 60} ,etc. Longest Increasing Subsequence: {10,22,33,50,60} => Length(LIS) = 5
- 3. Longest increasing subsequence Brute Force vs Dynamic Programming?
- 4. Longest increasing subsequence Dynamic Programming Approach: Iterator Array 10 22 9 33 21 50 41 60 LIS 1 1 1 1 1 1 1 1
- 5. Longest increasing subsequence Dynamic Programming Approach: For i = 1: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 1 1 1 1 1 1 1
- 6. Longest increasing subsequence Dynamic Programming Approach: For i = 1: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 1 1 1 1 1
- 7. Longest increasing subsequence Dynamic Programming Approach: For i = 2: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 1 1 1 1 1
- 8. Longest increasing subsequence Dynamic Programming Approach: For i = 3: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 1 1 1 1 1
- 9. Longest increasing subsequence Dynamic Programming Approach: For i = 3: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 2 1 1 1 1
- 10. Longest increasing subsequence Dynamic Programming Approach: For i = 3: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 3 1 1 1 1
- 11. Longest increasing subsequence Dynamic Programming Approach: For i = 4: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 3 1 1 1 1
- 12. Longest increasing subsequence Dynamic Programming Approach: For i = 4: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 3 2 1 1 1
- 13. Longest increasing subsequence Dynamic Programming Approach: For i = 4: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 3 2 1 1 1
- 14. Longest increasing subsequence Dynamic Programming Approach: For i = 4: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 3 2 1 1 1
- 15. Longest increasing subsequence Dynamic Programming Approach: For i = 5: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 3 2 1 1 1
- 16. Longest increasing subsequence Dynamic Programming Approach: For i = 5: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 3 2 2 1 1
- 17. Longest increasing subsequence Dynamic Programming Approach: For i = 5: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 3 2 3 1 1
- 18. Longest increasing subsequence Dynamic Programming Approach: For i = 5: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 3 2 3 1 1
- 19. Longest increasing subsequence Dynamic Programming Approach: For i = 5: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 3 2 4 1 1
- 20. Longest increasing subsequence Dynamic Programming Approach: Final Result: Iterator j i Array 10 22 9 33 21 50 41 60 LIS 1 2 1 3 2 4 4 5
- 21. Longest increasing subsequence Integer LongestIncreasingSubsequence( Integer[] array ) { Integer[] lis = new Integer[array.length]; for ( int i = 0 ; i < array.length ; i++) lis[i] = 1; for ( int i = 0 ; i < array.length ; i++) for ( int j = 0; j < i ; j++) if( array[j] < array [i] && lis[i] < lis[j] + 1) lis[i] = lis[j] + 1 return max( lis ) // O(n) }
Editor's Notes
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