给定两个数字序列P和Q。任务是找到两个序列的最长公共子序列，如果我们允许将第一个序列中最多k 个元素更改为任何值。
例子：
Input : P = { 8, 3 }
Q = { 1, 3 }
K = 1
Output : 2
If we change first element of first
sequence from 8 to 1, both sequences
become same.
Input : P = { 1, 2, 3, 4, 5 }
Q = { 5, 3, 1, 4, 2 }
K = 1
Output : 3
By changing first element of first
sequence to 5 to get the LCS ( 5, 3, 4 }.这个想法是使用动态规划。定义一个 3D 矩阵 dp[][][]，其中 dp[i][j][k] 定义第一个数组的前 i 个数字的最长公共子序列，第二个数组的第一个 j 个数字，当我们允许更改时第一个数组中的最大 k 数。
因此，递归看起来像
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)下面是这个方法的实现：
C++
// CPP program to find LCS of two arrays with
// k changes allowed in first array.
#include
using namespace std;
#define MAX 10
// Return LCS with at most k changes allowed.
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-1] == arr2[m-1])
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;
}
// Driven Program
int main()
{
int k = 1;
int arr1[] = { 1, 2, 3, 4, 5 };
int arr2[] = { 5, 3, 1, 4, 2 };
int n = sizeof(arr1) / sizeof(arr1[0]);
int m = sizeof(arr2) / sizeof(arr2[0]);
int dp[MAX][MAX][MAX];
memset(dp, -1, sizeof(dp));
cout << lcs(dp, arr1, n, arr2, m, k) << endl;
return 0;
} Java
// Java program to find LCS of two arrays with
// k changes allowed in first array.
class GFG
{
static int MAX = 10;
// Return LCS with at most k changes allowed.
static int lcs(int[][][] dp, int[] arr1,
int n, int[] arr2, int m, int k)
{
// If at most changes is less than 0.
if (k < 0)
return -10000000;
// 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;
try
{
// calculating LCS with no changes made.
ans = Math.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 - 1] == arr2[m - 1])
ans = Math.max(ans, 1 + lcs(dp, arr1, n - 1,
arr2, m - 1, k));
// calculating LCS with changes made.
ans = Math.max(ans, 1 + lcs(dp, arr1, n - 1,
arr2, m - 1, k - 1));
} catch (Exception e) { }
return ans;
}
// Driver Code
public static void main(String[] args)
{
int k = 1;
int[] arr1 = { 1, 2, 3, 4, 5 };
int[] arr2 = { 5, 3, 1, 4, 2 };
int n = arr1.length;
int m = arr2.length;
int[][][] dp = new int[MAX][MAX][MAX];
for (int i = 0; i < MAX; i++)
for (int j = 0; j < MAX; j++)
for (int l = 0; l < MAX; l++)
dp[i][j][l] = -1;
System.out.println(lcs(dp, arr1, n, arr2, m, k));
}
}
// This code is contributed by
// sanjeev2552Python3
# Python3 program to find LCS of two arrays
# with k changes allowed in the first array.
MAX = 10
# Return LCS with at most k changes allowed.
def lcs(dp, arr1, n, arr2, m, k):
# If at most changes is less than 0.
if k < 0:
return -(10 ** 7)
# If any of two array is over.
if n < 0 or m < 0:
return 0
# Making a reference variable to dp[n][m][k]
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-1] == arr2[m-1]:
ans = max(ans, 1 + lcs(dp, arr1, n - 1,
arr2, m - 1, k))
# calculating LCS with changes made.
ans = max(ans, lcs(dp, arr1, n - 1,
arr2, m - 1, k - 1))
return ans
# Driven Program
if __name__ == "__main__":
k = 1
arr1 = [1, 2, 3, 4, 5]
arr2 = [5, 3, 1, 4, 2]
n = len(arr1)
m = len(arr2)
dp = [[[-1 for i in range(MAX)] for j in range(MAX)] for k in range(MAX)]
print(lcs(dp, arr1, n, arr2, m, k))
# This code is contributed by Rituraj JainC#
// C# program to find LCS of two arrays with
// k changes allowed in first array.
using System;
class GFG
{
static int MAX = 10;
// Return LCS with at most
// k changes allowed.
static int lcs(int[,,] dp, int[] arr1,
int n, int[] arr2,
int m, int k)
{
// If at most changes is less than 0.
if (k < 0)
return -10000000;
// 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;
try
{
// calculating LCS with no changes made.
ans = Math.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 - 1] == arr2[m - 1])
ans = Math.Max(ans, 1 +
lcs(dp, arr1, n - 1,
arr2, m - 1, k));
// calculating LCS with changes made.
ans = Math.Max(ans, 1 +
lcs(dp, arr1, n - 1,
arr2, m - 1, k - 1));
} catch (Exception e) { }
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int k = 1;
int[] arr1 = { 1, 2, 3, 4, 5 };
int[] arr2 = { 5, 3, 1, 4, 2 };
int n = arr1.Length;
int m = arr2.Length;
int[,,] dp = new int[MAX, MAX, MAX];
for (int i = 0; i < MAX; i++)
for (int j = 0; j < MAX; j++)
for (int l = 0; l < MAX; l++)
dp[i, j, l] = -1;
Console.WriteLine(lcs(dp, arr1, n,
arr2, m, k));
}
}
// This code is contributed by PrinciRaj1992Javascript
输出：
3时间复杂度： O(N*M*K)。
如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。