给定大小为N的数组arr [] ，任务是计算三元组(arr [i]，arr [j]，arr [k])的数量，以使arr [i] * arr [j] = arr [j ] * arr [k] = 0 (i
Input: arr[] = {0, 8, 12, 0}
Output: 2
Explanation:
Triplets satisfying the given conditions are (0, 8, 0) and (0, 12, 0). Therefore, the required output is 2.
Input: arr[] = {1, 0, 2, 3}
Output: 2
天真的方法：解决此问题的最简单方法是从给定数组生成所有可能的三元组，并打印满足给定条件的三元组的计数。
时间复杂度： O(N 3 )
辅助空间： O(1)
高效方法：为了优化上述方法，其思想是使用Prefix Sum技术将0 s的计数存储在每个数组元素的左侧和右侧。通过将数组的当前元素视为arr [j]的值，遍历数组并计算满足给定条件的三元组的数量。最后，打印满足给定条件的三元组的计数。请按照以下步骤解决问题：
- 初始化一个数组，例如prefixZero [] ，为每个索引存储前面的索引中存在的0 s计数。
- 初始化变量TripletCnt以存储满足给定条件的三元组的数量。
- 遍历数组并检查arr [i]是否等于0 。如果发现为真，则将cntTriplet的值增加i *(N – i -1) 。
- 否则，将TripletCnt的值增加prefixZero [i] *(prefixZero [N – 1] – prefixZero [i]) 。
- 最后，打印TripletCnt的值。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to get the count
// of triples that satisfy
// the given condition
int cntTriplet(int arr[], int N)
{
// preZero[i] stores count
// of 0 up to index i
int preZero[N] = { 0 };
// Traverse the array and
// Count 0s up to index i
for (int i = 0; i < N; i++) {
if (arr[i] == 0) {
preZero[i]
= preZero[max(i - 1, 0)] + 1;
}
else {
preZero[i]
= preZero[max(i - 1, 0)];
}
}
// Stores count of triplet that
// satisfy the given conditions
int tripletCount = 0;
// Traverse the given array
for (int i = 0; i < N; i++) {
if (arr[i] == 0) {
// Stores count of elements
// on the left side of arr[i]
int X = i;
// Stores count of elements
// on the right side of arr[i]
int Y = N - i - 1;
tripletCount += X * Y;
}
else {
// Stores count of 0s on
// the left side of arr[i]
int X = preZero[i];
// Stores count of 0s on
// the right side of arr[i]
int Y = preZero[N - 1]
- preZero[i];
tripletCount += X * Y;
}
}
return tripletCount;
}
// Driver Code
int main()
{
int arr[] = { 1, 0, 2, 3 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << cntTriplet(arr, N);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to get the count
// of triples that satisfy
// the given condition
static int cntTriplet(int arr[],
int N)
{
// preZero[i] stores count
// of 0 up to index i
int []preZero = new int[N];
// Traverse the array and
// Count 0s up to index i
for (int i = 0; i < N; i++)
{
if (arr[i] == 0)
{
preZero[i] = preZero[Math.max(i - 1,
0)] + 1;
}
else
{
preZero[i] = preZero[Math.max(i - 1, 0)];
}
}
// Stores count of triplet that
// satisfy the given conditions
int tripletCount = 0;
// Traverse the given array
for (int i = 0; i < N; i++)
{
if (arr[i] == 0)
{
// Stores count of elements
// on the left side of arr[i]
int X = i;
// Stores count of elements
// on the right side of arr[i]
int Y = N - i - 1;
tripletCount += X * Y;
}
else
{
// Stores count of 0s on
// the left side of arr[i]
int X = preZero[i];
// Stores count of 0s on
// the right side of arr[i]
int Y = preZero[N - 1] -
preZero[i];
tripletCount += X * Y;
}
}
return tripletCount;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = {1, 0, 2, 3};
int N = arr.length;
System.out.print(cntTriplet(arr, N));
}
}
// This code contributed by gauravrajput1Python3
# Python3 program to implement
# the above appraoch
# Function to get the count
# of triples that satisfy
# the given condition
def cntTriplet(arr, N):
# preZero[i] stores count
# of 0 up to index i
preZero = [0] * N
# Traverse the array and
# Count 0s up to index i
for i in range(N):
if (arr[i] == 0):
preZero[i] = preZero[
max(i - 1, 0)] + 1
else:
preZero[i] = preZero[
max(i - 1, 0)]
# Stores count of triplet that
# satisfy the given conditions
tripletCount = 0
# Traverse the given array
for i in range(N):
if (arr[i] == 0):
# Stores count of elements
# on the left side of arr[i]
X = i
# Stores count of elements
# on the right side of arr[i]
Y = N - i - 1
tripletCount += X * Y
else:
# Stores count of 0s on
# the left side of arr[i]
X = preZero[i]
# Stores count of 0s on
# the right side of arr[i]
Y = preZero[N - 1] - preZero[i]
tripletCount += X * Y
return tripletCount
# Driver code
if __name__ == '__main__':
arr = [ 1, 0, 2, 3 ]
N = len(arr)
print(cntTriplet(arr, N))
# This code is contributed by Shivam SinghC#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to get the count
// of triples that satisfy
// the given condition
static int cntTriplet(int[] arr,
int N)
{
// preZero[i] stores count
// of 0 up to index i
int[] preZero = new int[N];
// Traverse the array and
// Count 0s up to index i
for (int i = 0; i < N; i++)
{
if (arr[i] == 0)
{
preZero[i] = preZero[Math.Max(i - 1,
0)] + 1;
}
else
{
preZero[i] = preZero[Math.Max(i - 1, 0)];
}
}
// Stores count of triplet that
// satisfy the given conditions
int tripletCount = 0;
// Traverse the given array
for (int i = 0; i < N; i++)
{
if (arr[i] == 0)
{
// Stores count of elements
// on the left side of arr[i]
int X = i;
// Stores count of elements
// on the right side of arr[i]
int Y = N - i - 1;
tripletCount += X * Y;
}
else
{
// Stores count of 0s on
// the left side of arr[i]
int X = preZero[i];
// Stores count of 0s on
// the right side of arr[i]
int Y = preZero[N - 1] -
preZero[i];
tripletCount += X * Y;
}
}
return tripletCount;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = {1, 0, 2, 3};
int N = arr.Length;
Console.Write(cntTriplet(arr, N));
}
}
// This code is contributed by Chitranayal输出：
2
时间复杂度： O(N)
辅助空间： O(N)