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📜  最长子序列的长度,以使相邻元素的xor不减小

📅  最后修改于: 2021-04-22 08:50:52             🧑  作者: Mango

给定N个正整数的序列arr ,任务是找到最长子序列的长度,以使子序列中相邻整数的xor必须不减

例子:

方法:

  • 使用动态编程可以解决此问题,其中dp [i]将存储以索引i结尾的最长有效子序列的长度。
  • 首先,存储所有元素对(即arr [i] ^ arr [j])元素对(i,j)的xor,然后根据xor的值对它们进行排序,因为它们需要不减小。
  • 现在,如果对(i,j)的被认为是随后的最长子序列的长度与j的末端将是最大(DP [J],1 + DP [I])。这样,为每个位置计算dp []数组的最大可能值,然后取其中的最大值。

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
 
// Function to find the length of the longest
// subsequence such that the XOR of adjacent
// elements in the subsequence must
// be non-decreasing
int LongestXorSubsequence(int arr[], int n)
{
 
    vector > > v;
 
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // Computing xor of all the pairs
            // of elements and store them
            // along with the pair (i, j)
            v.push_back(make_pair(arr[i] ^ arr[j],
                                  make_pair(i, j)));
        }
    }
 
    // Sort all possible xor values
    sort(v.begin(), v.end());
 
    int dp[n];
 
    // Initialize the dp array
    for (int i = 0; i < n; i++) {
        dp[i] = 1;
    }
 
    // Calculating the dp array
    // for each possible position
    // and calculating the max length
    // that ends at a particular index
    for (auto i : v) {
        dp[i.second.second]
            = max(dp[i.second.second],
                  1 + dp[i.second.first]);
    }
 
    int ans = 1;
 
    // Taking maximum of all position
    for (int i = 0; i < n; i++)
        ans = max(ans, dp[i]);
 
    return ans;
}
 
// Driver code
int main()
{
 
    int arr[] = { 2, 12, 6, 7, 13, 14, 8, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << LongestXorSubsequence(arr, n);
 
    return 0;
}


Python3
# Python3 implementation of the approach
 
# Function to find the length of the longest
# subsequence such that the XOR of adjacent
# elements in the subsequence must
# be non-decreasing
def LongestXorSubsequence(arr, n):
 
    v = []
 
    for i in range(0, n):
        for j in range(i + 1, n):
 
             # Computing xor of all the pairs
            # of elements and store them
            # along with the pair (i, j)
            v.append([(arr[i] ^ arr[j]), (i, j)])
 
        # v.push_back(make_pair(arr[i] ^ arr[j], make_pair(i, j)))
         
    # Sort all possible xor values
    v.sort()
     
    # Initialize the dp array
    dp = [1 for x in range(88)]
 
    # Calculating the dp array
    # for each possible position
    # and calculating the max length
    # that ends at a particular index
    for a, b in v:
        dp[b[1]] = max(dp[b[1]], 1 + dp[b[0]])
     
    ans = 1
 
    # Taking maximum of all position
    for i in range(0, n):
        ans = max(ans, dp[i])
 
    return ans
 
# Driver code
arr = [ 2, 12, 6, 7, 13, 14, 8, 6 ]
n = len(arr)
print(LongestXorSubsequence(arr, n))
 
# This code is contributed by Sanjit Prasad


输出
5

时间复杂度: O(N * N)
辅助空间: O(N)

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