LintCode 762 - Longest Common Subsequence II


Related: Longest Common Subsequence
https://www.lintcode.com/problem/longest-common-subsequence-ii/description
Given two sequence P and Q of numbers. The task is to find Longest Common Subsequence of two sequence if we are allowed to change at most k element in first sequence to any value.

Example

Given P = [8 ,3], Q = [1, 3], K = 1
Return 2
Given P = [1, 2, 3, 4, 5], Q = [5, 3, 1, 4, 2], K = 1
Return 3
https://www.geeksforgeeks.org/longest-common-subsequence-with-at-most-k-changes-allowed/
The idea is to use Dynamic Programming. Define a 3D matrix dp[][][], where dp[i][j][k] defines the Longest Common Subsequence for the first i numbers of first array, first j number of second array when we are allowed to change at max k number in the first array.
Therefore, recursion will look like
If P[i] != Q[j], 
    dp[i][j][k] = max(dp[i - 1][j][k], 
                      dp[i][j - 1][k], 
                      dp[i - 1][j - 1][k - 1] + 1) 
If P[i] == Q[j], 
    dp[i][j][k] = max(dp[i - 1][j][k], 
                      dp[i][j - 1][k], 
                      dp[i - 1][j - 1][k] + 1)
Time Complexity: O(N*M*K).

int lcs(int dp[MAX][MAX][MAX], int arr1[], int n,
                       int arr2[], int m, int k)
{
    // If at most changes is less than 0.
    if (k < 0)
        return -1e7;
  
    // If any of two array is over.
    if (n < 0 || m < 0)
        return 0;
  
    // Making a reference variable to dp[n][m][k]
    int& ans = dp[n][m][k];
  
    // If value is already calculated, return
    // that value.
    if (ans != -1)
        return ans;
  
    // calculating LCS with no changes made.
    ans = max(lcs(dp, arr1, n - 1, arr2, m, k), 
              lcs(dp, arr1, n, arr2, m - 1, k));
  
    // calculating LCS when array element are same.
    if (arr1[n] == arr2[m])
        ans = max(ans, 1 + lcs(dp, arr1, n - 1, 
                                arr2, m - 1, k));
  
    // calculating LCS with changes made.
    ans = max(ans, 1 + lcs(dp, arr1, n - 1, 
                          arr2, m - 1, k - 1));
  
    return ans;
}

https://www.jiuzhang.com/solution/longest-common-subsequence-ii/
最长公共子串的问题是典型的动态规划题,在这里已有题解,就是用二维矩阵来存储所有的中间最佳状态。
这是这个题的变种,因为第一个字符串可以进行k次变动。那么解体就得用三维矩阵来存储中间最佳状态。
而且要把k=0的情况拿出来单独解决。这个k=0的情况实际上就是无变动情形下的解。这个的时间复杂度
是O(mnk),m和n是P和Q字符串的长度,k是P可变换元素数量。
    def longestCommonSubsequence2(self, P, Q, k):
        # Write your code here
        m = len(P)
        n = len(Q)
        l = [[[None for g in range(k+1)] \
        for j in range(n+1)] for i in range(m+1)]
        for i in range(m+1):
            for j in range(n+1):
                if i == 0 or j == 0:
                    l[i][j][0] = 0
                elif P[i-1] == Q[j-1]:
                    l[i][j][0] = l[i-1][j-1][0] + 1
                else:
                    l[i][j][0] = max(l[i-1][j][0], l[i][j-1][0])
        for i in range(m+1):
            for j in range(n+1):
                for g in range(1, k+1):
                    if i == 0  or j == 0:
                        l[i][j][g] = 0
                    elif P[i-1] != Q[j-1]:
                        l[i][j][g] = max(l[i-1][j][g], l[i][j-1][g], l[i-1][j-1][g-1] + 1)
                    else:
                        l[i][j][g] = max(l[i-1][j][g], l[i][j-1][g], l[i-1][j-1][g] + 1)
        return l[m][n][k]

    https://blog.csdn.net/zhaohengchuan/article/details/79703980
    dp[i][j][t]表示P的前i个数字和Q的前j个数字中,在变换了t个元素的情况下,最长公共子序列的长度。初始化之后,对于dp[i][j][t],如果P[i-1]=Q[j-1],由于t表示改变t个元素,dp[i][j][t]可能取dp[i-1][j-1][t]+1,在P[i-1]!=Q[j-1]时,dp[i][j][t]可能取dp[i-1][j-1][t-1]+1(需要改变一个元素)。由于dp[i][j][t]还可能取dp[i-1][j][t]和dp[i][j-1][t],最终结果取最大值。
        int longestCommonSubsequence2(vector<int> &P, vector<int> &Q, int k) {
            // Write your code here
            if (P.empty() || Q.empty() || k < 0)
                return 0;
            int lenp = P.size(), lenq = Q.size();
            //dp[i][j][t]表示P的前i个数字和Q的前j个数字中,在变换了t个元素的情况下,最长公共子序列的长度
            vector<vector<vector<int>>> dp(lenp + 1, vector<vector<int>>(lenq + 1, vector<int>(k + 1)));
            //当i = 0或j = 0时,dp[i][j][t] = 0
            for (int i = 0; i <= lenp; ++i)
            {
                for (int t = 0; t <= k; ++t)
                    dp[i][0][t] = 0;
            }
            for (int j = 0; j <= lenq; ++j)
            {
                for (int t = 0; t <= k; ++t)
                    dp[0][j][t] = 0;
            }
            //处理k=0的情况,类似于公共子序列
            for (int i = 1; i <= lenp; ++i)
            {
                for (int j = 1; j <= lenq; ++j)
                {
                    if (P[i - 1] == Q[j - 1])
                        dp[i][j][0] = dp[i - 1][j - 1][0] + 1;
                    else
                        dp[i][j][0] = maxVal(dp[i - 1][j][0], dp[i][j - 1][0]);
                }
            }
            //若P[i-1]=Q[j-1],不必再变换元素,dp[i][j][t]是dp[i-1][j][t],dp[i][j-1][t],dp[i-1][j-1][t]+1的最大值
            //否则,可能需要变换元素,dp[i][j][t]是dp[i-1][j][t],dp[i][j-1][t],dp[i-1][j-1][t-1]+1的最大值
            for (int i = 1; i <= lenp; ++i)
            {
                for (int j = 1; j <= lenq; ++j)
                {
                    for (int t = 1; t <= k; ++t)
                    {
                        if (P[i - 1] == Q[j - 1])
                            dp[i][j][t] = maxVal(dp[i - 1][j][t], maxVal(dp[i][j - 1][t], dp[i - 1][j - 1][t] + 1));
                        else
                            dp[i][j][t] = maxVal(dp[i - 1][j][t], maxVal(dp[i][j - 1][t], dp[i - 1][j - 1][t - 1] + 1));
                    }
                }
            }
            return dp[lenp][lenq][k];
        }
    https://github.com/jiadaizhao/LintCode/blob/master/0762-Longest%20Common%20Subsequence%20II/0762-Longest%20Common%20Subsequence%20II.cpp
        int longestCommonSubsequence2(vector<int> &P, vector<int> &Q, int k) {
            // Write your code here
            int m = P.size();
            int n = Q.size();
            int dp[1 + k][1 + m][1 + n];
            memset(dp, 0, sizeof(dp));
            for (int i = 1; i <= m; ++i) {
                for (int j = 1; j <= n; ++j) {
                    if (P[i - 1] == Q[j - 1]) {
                        dp[0][i][j] = 1 + dp[0][i - 1][j - 1];
                    }
                    else {
                        dp[0][i][j] = max(dp[0][i - 1][j], dp[0][i][j - 1]);
                    }
                }
            }
         
            for (int l = 1; l <= k; ++l) {
                for (int i = 1; i <= m; ++i) {
                    for (int j = 1; j <= n; ++j) {
                        if (P[i - 1] == Q[j - 1]) {
                            dp[l][i][j] = 1 + dp[l][i - 1][j - 1];
                        }
                        else {
                            dp[l][i][j] = max({dp[l][i - 1][j], dp[l][i][j - 1], 1 + dp[l - 1][i - 1][j - 1]});
                        }
                    }
                }
            }
         
            return dp[k][m][n];
        }




    Labels

    LeetCode (1432) GeeksforGeeks (1122) LeetCode - Review (1067) Review (882) Algorithm (668) to-do (609) Classic Algorithm (270) Google Interview (237) Classic Interview (222) Dynamic Programming (220) DP (186) Bit Algorithms (145) POJ (141) Math (137) Tree (132) LeetCode - Phone (129) EPI (122) Cracking Coding Interview (119) DFS (115) Difficult Algorithm (115) Lintcode (115) Different Solutions (110) Smart Algorithm (104) Binary Search (96) BFS (91) HackerRank (90) Binary Tree (86) Hard (79) Two Pointers (78) Stack (76) Company-Facebook (75) BST (72) Graph Algorithm (72) Time Complexity (69) Greedy Algorithm (68) Interval (63) Company - Google (62) Geometry Algorithm (61) Interview Corner (61) LeetCode - Extended (61) Union-Find (60) Trie (58) Advanced Data Structure (56) List (56) Priority Queue (53) Codility (52) ComProGuide (50) LeetCode Hard (50) Matrix (50) Bisection (48) Segment Tree (48) Sliding Window (48) USACO (46) Space Optimization (45) Company-Airbnb (41) Greedy (41) Mathematical Algorithm (41) Tree - Post-Order (41) ACM-ICPC (40) Algorithm Interview (40) Data Structure Design (40) Graph (40) Backtracking (39) Data Structure (39) Jobdu (39) Random (39) Codeforces (38) Knapsack (38) LeetCode - DP (38) Recursive Algorithm (38) String Algorithm (38) TopCoder (38) Sort (37) Introduction to Algorithms (36) Pre-Sort (36) Beauty of Programming (35) Must Known (34) Binary Search Tree (33) Follow Up (33) prismoskills (33) Palindrome (32) Permutation (31) Array (30) Google Code Jam (30) HDU (30) Array O(N) (29) Logic Thinking (29) Monotonic Stack (29) Puzzles (29) Code - Detail (27) Company-Zenefits (27) Microsoft 100 - July (27) Queue (27) Binary Indexed Trees (26) TreeMap (26) to-do-must (26) 1point3acres (25) GeeksQuiz (25) Merge Sort (25) Reverse Thinking (25) hihocoder (25) Company - LinkedIn (24) Hash (24) High Frequency (24) Summary (24) Divide and Conquer (23) Proof (23) Game Theory (22) Topological Sort (22) Lintcode - Review (21) Tree - Modification (21) Algorithm Game (20) CareerCup (20) Company - Twitter (20) DFS + Review (20) DP - Relation (20) Brain Teaser (19) DP - Tree (19) Left and Right Array (19) O(N) (19) Sweep Line (19) UVA (19) DP - Bit Masking (18) LeetCode - Thinking (18) KMP (17) LeetCode - TODO (17) Probabilities (17) Simulation (17) String Search (17) Codercareer (16) Company-Uber (16) Iterator (16) Number (16) O(1) Space (16) Shortest Path (16) itint5 (16) DFS+Cache (15) Dijkstra (15) Euclidean GCD (15) Heap (15) LeetCode - Hard (15) Majority (15) Number Theory (15) Rolling Hash (15) Tree Traversal (15) Brute Force (14) Bucket Sort (14) DP - Knapsack (14) DP - Probability (14) Difficult (14) Fast Power Algorithm (14) Pattern (14) Prefix Sum (14) TreeSet (14) Algorithm Videos (13) Amazon Interview (13) Basic Algorithm (13) Codechef (13) Combination (13) Computational Geometry (13) DP - Digit (13) LCA (13) LeetCode - DFS (13) Linked List (13) Long Increasing Sequence(LIS) (13) Math-Divisible (13) Reservoir Sampling (13) mitbbs (13) Algorithm - How To (12) Company - Microsoft (12) DP - Interval (12) DP - Multiple Relation (12) DP - Relation Optimization (12) LeetCode - Classic (12) Level Order Traversal (12) Prime (12) Pruning (12) Reconstruct Tree (12) Thinking (12) X Sum (12) AOJ (11) Bit Mask (11) Company-Snapchat (11) DP - Space Optimization (11) Dequeue (11) Graph DFS (11) MinMax (11) Miscs (11) Princeton (11) Quick Sort (11) Stack - Tree (11) 尺取法 (11) 挑战程序设计竞赛 (11) Coin Change (10) DFS+Backtracking (10) Facebook Hacker Cup (10) Fast Slow Pointers (10) HackerRank Easy (10) Interval Tree (10) Limited Range (10) Matrix - Traverse (10) Monotone Queue (10) SPOJ (10) Starting Point (10) States (10) Stock (10) Theory (10) Tutorialhorizon (10) Kadane - Extended (9) Mathblog (9) Max-Min Flow (9) Maze (9) Median (9) O(32N) (9) Quick Select (9) Stack Overflow (9) System Design (9) Tree - Conversion (9) Use XOR (9) Book Notes (8) Company-Amazon (8) DFS+BFS (8) DP - States (8) Expression (8) Longest Common Subsequence(LCS) (8) One Pass (8) Quadtrees (8) Traversal Once (8) Trie - Suffix (8) 穷竭搜索 (8) Algorithm Problem List (7) All Sub (7) Catalan Number (7) Cycle (7) DP - Cases (7) Facebook Interview (7) Fibonacci Numbers (7) Flood fill (7) Game Nim (7) Graph BFS (7) HackerRank Difficult (7) Hackerearth (7) Inversion (7) Kadane’s Algorithm (7) Manacher (7) Morris Traversal (7) Multiple Data Structures (7) Normalized Key (7) O(XN) (7) Radix Sort (7) Recursion (7) Sampling (7) Suffix Array (7) Tech-Queries (7) Tree - Serialization (7) Tree DP (7) Trie - Bit (7) 蓝桥杯 (7) Algorithm - Brain Teaser (6) BFS - Priority Queue (6) BFS - Unusual (6) Classic Data Structure Impl (6) DP - 2D (6) DP - Monotone Queue (6) DP - Unusual (6) DP-Space Optimization (6) Dutch Flag (6) How To (6) Interviewstreet (6) Knapsack - MultiplePack (6) Local MinMax (6) MST (6) Minimum Spanning Tree (6) Number - Reach (6) Parentheses (6) Pre-Sum (6) Probability (6) Programming Pearls (6) Rabin-Karp (6) Reverse (6) Scan from right (6) Schedule (6) Stream (6) Subset Sum (6) TSP (6) Xpost (6) n00tc0d3r (6) reddit (6) AI (5) Abbreviation (5) Anagram (5) Art Of Programming-July (5) Assumption (5) Bellman Ford (5) Big Data (5) Code - Solid (5) Code Kata (5) Codility-lessons (5) Coding (5) Company - WMware (5) Convex Hull (5) Crazyforcode (5) DFS - Multiple (5) DFS+DP (5) DP - Multi-Dimension (5) DP-Multiple Relation (5) Eulerian Cycle (5) Graph - Unusual (5) Graph Cycle (5) Hash Strategy (5) Immutability (5) Java (5) LogN (5) Manhattan Distance (5) Matrix Chain Multiplication (5) N Queens (5) Pre-Sort: Index (5) Quick Partition (5) Quora (5) Randomized Algorithms (5) Resources (5) Robot (5) SPFA(Shortest Path Faster Algorithm) (5) Shuffle (5) Sieve of Eratosthenes (5) Strongly Connected Components (5) Subarray Sum (5) Sudoku (5) Suffix Tree (5) Swap (5) Threaded (5) Tree - Creation (5) Warshall Floyd (5) Word Search (5) jiuzhang (5)

    Popular Posts