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📜  最长子数组,以使max和min之差最大为K

📅  最后修改于: 2021-04-29 17:09:14             🧑  作者: Mango

给定长度为N的数组arr [] ,tark将找到最长子序列的长度,以使其最大元素和最小元素的差不大于整数K。

如果ααa可以通过删除几个(可能为零)元素从b中获得,则序列a是序列b的子序列。例如, [3,1] [3,1][3,2,1] [3,2,1][4,3,1] [4,3,1]的子序列,但不是[1,3,3,7] [1,3,3,7][3,10,4] [3,10,4]的亚序列
例子:

天真的方法:

  • 生成所有子数组,并在子数组中找到最小值和最大值。
  • 计算最小值和最大值之间的差异,如果该差异小于或等于K,则更新答案

时间复杂度: O(N 3 )
高效的方法:想法是首先对数组进行排序,然后使用二进制搜索来优化方法。

  • 对给定数组进行排序
  • 对于数组中每个不同的元素A [i] ,找到第一个元素A [j]使得(A [j] -A [i])> K。
  • 对于它的实现,我们使用Binary searchlower_bound并每次更新ans作为先前ans的最大值和indixes的差。

下面是上述方法的实现:

CPP
// C++ program to find longest
// subarray such that difference
// of max and min is at-most K
 
#include 
using namespace std;
 
// Function to calculate longest
// subarray with above condition
int findLargestSubarray(
    vector& arr,
    int N, int K)
{
 
    // Sort the array
    sort(arr.begin(), arr.end());
 
    int value1 = arr[0], value2 = 0;
    int index1, index2, i, MAX;
    index1 = index2 = 0;
    i = 0, MAX = 0;
 
    // Loop which will terminate
    // when no further elements
    // can be included in the subarray
 
    while (index2 != N) {
 
        // first value such that
        // arr[index2] - arr[index1] > K
        value2 = value1 + (K + 1);
 
        // calculate its index using lower_bound
        index2 = lower_bound(arr.begin(),
                             arr.end(), value2)
                 - arr.begin();
 
        // index2- index1 will give
        // the accurate length
        // of suarray then compare
        // for MAX length and store
        // in MAX variable
        MAX = max(MAX, (index2 - index1));
 
        // change the index1
        // to next greater element
        // than previous one
        // and recalculate the value1
        index1
            = lower_bound(
                  arr.begin(),
                  arr.end(), arr[index1] + 1)
              - arr.begin();
        value1 = arr[index1];
    }
 
    // finally return answer MAX
    return MAX;
}
// Driver Code
int main()
{
    int N, K;
    N = 18;
    K = 5;
    vector arr{ 1, 1, 1, 2, 2,
                     2, 2, 2, 3,
                     3, 3, 6, 6, 7,
                     7, 7, 7, 7 };
    cout << findLargestSubarray(arr, N, K);
    return 0;
}


输出:
15

时间复杂度: O(N * log(N))