给定的阵列ARR大小为N [],和整数K,任务是找到数组元素的计数的常用3 [I],其是大于(ARR [I – 1] + ARR [I – 2] + … + arr [i – K]) 。
例子:
Input: arr[] = { 2, 3, 8, 10, -2, 7, 5, 5, 9, 15 }, K = 2
Output: 2
Explanation:
arr[2]( > arr[1] + arr[0]) and arr[9]( > arr[8] + arr[7]) are the two array elements exceeding sum of preceding K( = 2) elements.
Input: arr[] = { 17, -2, 16, -8, 19, 11, 5, 15, -9, 24 }, K = 3
Output: 2
方法:请按照以下步骤解决问题:
- 计算并存储给定数组的前缀和。
- 遍历索引[K,N – 1]中存在的数组元素。
- 对于每个数组元素,检查arr [i]是否超过(arr [i – 1] + arr [i – 2] +…+ arr [i – K]) 。因此,任务是检查k> i> N是否为arr [i]超过前缀[i – 1] –前缀[i –(K + 1)] ,或者是否为arr [i]超过前缀[i – 1] i = K
- 满足上述条件的数组元素的增量计数。最后,打印所需的计数。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count array elements
// exceeding sum of preceding K elements
int countPrecedingK(int a[], int n, int K)
{
int prefix[n];
prefix[0] = a[0];
// Iterate over the array
for (int i = 1; i < n; i++) {
// Update prefix sum
prefix[i]
= prefix[i - 1] + a[i];
}
int ctr = 0;
// Check if arr[K] >
// arr[0] + .. + arr[K - 1]
if (prefix[K - 1] < a[K])
// Increment count
ctr++;
for (int i = K + 1; i < n; i++) {
// Check if arr[i] >
// arr[i - K - 1] + .. + arr[i - 1]
if (prefix[i - 1] - prefix[i - K - 1]
< a[i])
// Increment count
ctr++;
}
return ctr;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 2, 3, 8, 10, -2,
7, 5, 5, 9, 15 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 2;
cout << countPrecedingK(arr, N, K);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to count array elements
// exceeding sum of preceding K elements
static int countPrecedingK(int a[], int n,
int K)
{
int []prefix = new int[n];
prefix[0] = a[0];
// Iterate over the array
for(int i = 1; i < n; i++)
{
// Update prefix sum
prefix[i] = prefix[i - 1] + a[i];
}
int ctr = 0;
// Check if arr[K] >
// arr[0] + .. + arr[K - 1]
if (prefix[K - 1] < a[K])
// Increment count
ctr++;
for(int i = K + 1; i < n; i++)
{
// Check if arr[i] >
// arr[i - K - 1] + .. + arr[i - 1]
if (prefix[i - 1] -
prefix[i - K - 1] < a[i])
// Increment count
ctr++;
}
return ctr;
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 2, 3, 8, 10, -2,
7, 5, 5, 9, 15 };
int N = arr.length;
int K = 2;
System.out.print(countPrecedingK(arr, N, K));
}
}
// This code is contributed by Amit Katiyar
Python3
# Python3 program for the above approach
# Function to count array elements
# exceeding sum of preceding K elements
def countPrecedingK(a, n, K):
prefix = [0] * n
prefix[0] = a[0]
# Iterate over the array
for i in range(1, n):
# Update prefix sum
prefix[i] = prefix[i - 1] + a[i]
ctr = 0
# Check if arr[K] >
# arr[0] + .. + arr[K - 1]
if (prefix[K - 1] < a[K]):
# Increment count
ctr += 1
for i in range(K + 1, n):
# Check if arr[i] >
# arr[i - K - 1] + .. + arr[i - 1]
if (prefix[i - 1] -
prefix[i - K - 1] < a[i]):
# Increment count
ctr += 1
return ctr
# Driver Code
if __name__ == '__main__':
# Given array
arr = [ 2, 3, 8, 10, -2,
7, 5, 5, 9, 15 ]
N = len(arr)
K = 2
print(countPrecedingK(arr, N, K))
# This code is contributed by mohit kumar 29
C#
// C# program for
// the above approach
using System;
class GFG{
// Function to count array elements
// exceeding sum of preceding K elements
static int countPrecedingK(int []a,
int n, int K)
{
int []prefix = new int[n];
prefix[0] = a[0];
// Iterate over the array
for(int i = 1; i < n; i++)
{
// Update prefix sum
prefix[i] = prefix[i - 1] + a[i];
}
int ctr = 0;
// Check if arr[K] >
// arr[0] + .. + arr[K - 1]
if (prefix[K - 1] < a[K])
// Increment count
ctr++;
for(int i = K + 1; i < n; i++)
{
// Check if arr[i] >
// arr[i - K - 1] + .. + arr[i - 1]
if (prefix[i - 1] -
prefix[i - K - 1] < a[i])
// Increment count
ctr++;
}
return ctr;
}
// Driver Code
public static void Main(String[] args)
{
// Given array
int []arr = {2, 3, 8, 10, -2,
7, 5, 5, 9, 15};
int N = arr.Length;
int K = 2;
Console.Write(countPrecedingK(arr, N, K));
}
}
// This code is contributed by Princi Singh
输出:
2
时间复杂度: O(N)
辅助空间: O(N)