Number of subarrays with given product
Given an array of positive numbers and a number k, find the number of subarrays having product exactly equal to k. We may assume that there is no overflow.
Examples :
Input : arr = [2, 1, 1, 1, 4, 5]
k = 4
Output : 4
1st subarray : arr[1..4]
2nd subarray : arr[2..4]
3rd subarray : arr[3..4]
4th subarray : arr[4]
Input : arr = [1, 2, 3, 4, 1]
k = 24
Output : 4
1st subarray : arr[0..4]
2nd subarray : arr[1..4]
3rd subarray : arr[1..3]
4th subarray : arr[0..3]
A simple solution is to consider all subarrays and find their products. For every product, check if it is equal to k. If yes, then increment result.
An efficient solution is to use sliding window technique can be used to solve the problem. We use two pointers start and end to represent starting and ending point of sliding window.
Initially both start and end point to the beginning of the array, i.e. index 0. Keep incrementing end until product p < k. As soon as p becomes equal to k stop incrementing end and check whether the current position of end is followed by a series of 1s in the array. If yes then those 1s will also contribute to the subarray count. Store the count of these succeeding 1s in a variable countOnes. After this keep incrementing start until p equals to k while adding (countOnes + 1) to result. If start coincides with end then again start from incrementing end and follow the same procedure. Do this until end < array size.
Why countOnes + 1 is added to result?
Consider the second test case in the above sample cases. If we follow the above mentioned procedure, then after incrementing end, we will reach at start = 0 and end = 3. After this the countOnes is set equal to 1. How many subarrays are there for start = 0 ? There are two subarrays : arr[0..3] and arr[0..4]. Observe that subarray[0..3] is what we have found using sliding window technique. This increases the result count by 1 and is what represented by + 1 in expression countOnes + 1. The other subarray[0..4] is simply obtained by appending single 1 to the subarray[0..3], i.e. adding countOnes number of 1s one at a time. Let us try to generalise it. Suppose arr[0..i] is the subarray obtained using sliding window technique and let countOnes = j. Then we can expand this subarray by unit length at a time by appending single 1 to this subarray. After appending a single 1 to arr[0..i] the new subarray is arr[0..i+1] and result is also increased by 1. countOnes now decrease by 1 and equals j – 1. We can continuously append single 1 at a time and obtain a new subarray until countOnes is not equal to zero.
Thus the result count increases by countOnes and is represented as countOnes in expression countOnes + 1. So for every increment in start until p equals k simply add countOnes + 1 to the result.
Note that above algorithm will not work for the case when k = 1. e.g. Consider the test case arr[] = {2, 1, 1, 1}. Thanks to Jeel Santoki for this test case. For the case when k = 1, we will find length of each segment of array in which all elements are 1. Let the length of particular segment of 1 is x. The number of subarrays of this segment will be x*(x+1)/2. All these subarrays will have product 1 as all elements are 1. In given test case there is only one segment of 1 from index 1 to index 3 having length 3. So total subarrays with product 1 are (3*4)/2 = 6.
The algorithm can be listed as:
For k != 1:
1. Initialize start = end = 0
2. Initialize res = 0, p = 1
3. Increment end until p < k
4. When p = k do:
Set countOnes = number of succeeding ones
res += (countOnes+1)
Increment start until p = k
and do res += (countOnes+1)
5. Stop if end = n
For k = 1:
1. Find all segments in array in which
only 1 is present.
2. Find length of each segment.
3. Add length*(length+1) / 2 to result.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int countOne( int arr[], int n){
int i = 0;
int len = 0;
int ans = 0;
while (i < n){
if (arr[i] == 1){
len = 0;
while (i < n && arr[i] == 1){
i++;
len++;
}
ans += (len*(len+1)) / 2;
}
i++;
}
return ans;
}
int findSubarrayCount( int arr[], int n, int k)
{
int start = 0, endval = 0, p = 1,
countOnes = 0, res = 0;
while (endval < n)
{
p *= (arr[endval]);
if (p > k)
{
while (start <= endval && p > k)
{
p /= arr[start];
start++;
}
}
if (p == k)
{
countOnes = 0;
while (endval + 1 < n && arr[endval + 1] == 1)
{
countOnes++;
endval++;
}
res += (countOnes + 1);
while (start <= endval && arr[start] == 1 && k!=1)
{
res += (countOnes + 1);
start++;
}
p /= arr[start];
start++;
}
endval++;
}
return res;
}
int main()
{
int arr1[] = { 2, 1, 1, 1, 3, 1, 1, 4};
int n1 = sizeof (arr1) / sizeof (arr1[0]);
int k = 1;
if (k != 1)
cout << findSubarrayCount(arr1, n1, k) << "\n" ;
else
cout << countOne(arr1, n1) << "\n" ;
int arr2[] = { 2, 1, 1, 1, 4, 5};
int n2 = sizeof (arr2) / sizeof (arr2[0]);
k = 4;
if (k != 1)
cout << findSubarrayCount(arr2, n2, k) << "\n" ;
else
cout << countOne(arr2, n2) << "\n" ;
return 0;
}
|
Java
import java.util.*;
class GFG
{
public static int findSubarrayCount( int arr[], int n, int k)
{
int start = 0 , endval = 0 ;
int p = 1 , countOnes = 0 , res = 0 ;
while (endval < n)
{
p *= (arr[endval]);
if (p > k)
{
while (start <= endval && p > k)
{
p /= arr[start];
start++;
}
}
if (p == k)
{
countOnes = 0 ;
while (endval + 1 < n && arr[endval + 1 ] == 1 )
{
countOnes++;
endval++;
}
res += (countOnes + 1 );
while (start <= endval && arr[start] == 1 )
{
res += (countOnes + 1 );
start++;
}
p /= arr[start];
start++;
}
endval++;
}
return res;
}
public static void main (String[] args)
{
int arr[] = new int []{ 2 , 1 , 1 , 1 , 4 , 5 };
int n = arr.length;
int k = 4 ;
System.out.println(findSubarrayCount(arr, n, k));
}
}
|
Python3
def countOne(arr, n) :
i = 0
Len = 0
ans = 0
while (i < n) :
if (arr[i] = = 1 ) :
Len = 0
while (i < n and arr[i] = = 1 ) :
i + = 1
Len + = 1
ans + = ( Len * ( Len + 1 )) / / 2
i + = 1
return ans
def findSubarrayCount(arr, n, k) :
start, endval, p, countOnes, res = 0 , 0 , 1 , 0 , 0
while (endval < n) :
p = p * (arr[endval])
if (p > k) :
while (start < = endval and p > k) :
p = p / / arr[start]
start + = 1
if (p = = k) :
countOnes = 0
while endval + 1 < n and arr[endval + 1 ] = = 1 :
countOnes + = 1
endval + = 1
res + = (countOnes + 1 )
while (start < = endval and arr[start] = = 1 and k! = 1 ) :
res + = (countOnes + 1 )
start + = 1
p = p / / arr[start]
start + = 1
endval + = 1
return res
arr1 = [ 2 , 1 , 1 , 1 , 3 , 1 , 1 , 4 ]
n1 = len (arr1)
k = 1
if (k ! = 1 ) :
print (findSubarrayCount(arr1, n1, k))
else :
print (countOne(arr1, n1))
arr2 = [ 2 , 1 , 1 , 1 , 4 , 5 ]
n2 = len (arr2)
k = 4
if (k ! = 1 ) :
print (findSubarrayCount(arr2, n2, k))
else :
print (countOne(arr2, n2))
|
C#
using System;
class GFG
{
public static int findSubarrayCount( int []arr,
int n, int k)
{
int start = 0, endval = 0;
int p = 1, countOnes = 0, res = 0;
while (endval < n)
{
p *= (arr[endval]);
if (p > k)
{
while (start <= endval && p > k)
{
p /= arr[start];
start++;
}
}
if (p == k)
{
countOnes = 0;
while (endval + 1 < n &&
arr[endval + 1] == 1)
{
countOnes++;
endval++;
}
res += (countOnes + 1);
while (start <= endval &&
arr[start] == 1)
{
res += (countOnes + 1);
start++;
}
p /= arr[start];
start++;
}
endval++;
}
return res;
}
public static void Main ()
{
int []arr = new int []{ 2, 1, 1,
1, 4, 5 };
int n = arr.Length;
int k = 4;
Console.WriteLine(findSubarrayCount(arr, n, k));
}
}
|
PHP
<?php
function findSubarrayCount( $arr , $n , $k )
{
$start = 0;
$endval = 0;
$p = 1;
$countOnes = 0;
$res = 0;
while ( $endval < $n )
{
$p *= ( $arr [ $endval ]);
if ( $p > $k )
{
while ( $start <= $endval && $p > $k )
{
$p /= $arr [ $start ];
$start ++;
}
}
if ( $p == $k )
{
$countOnes = 0;
while ( $endval + 1 < $n &&
$arr [ $endval + 1] == 1)
{
$countOnes ++;
$endval ++;
}
$res += ( $countOnes + 1);
while ( $start <= $endval &&
$arr [ $start ] == 1)
{
$res += ( $countOnes + 1);
$start ++;
}
$p /= $arr [ $start ];
$start ++;
}
$endval ++;
}
return $res ;
}
$arr = array (2, 1, 1, 1, 4, 5);
$n = sizeof( $arr ) ;
$k = 4;
echo findSubarrayCount( $arr , $n , $k );
?>
|
Javascript
<script>
function countOne(arr, n)
{
let i = 0;
let len = 0;
let ans = 0;
while (i < n)
{
if (arr[i] == 1)
{
len = 0;
while (i < n && arr[i] == 1)
{
i++;
len++;
}
ans += parseInt((len * (len + 1)) / 2, 10);
}
i++;
}
return ans;
}
function findSubarrayCount(arr, n, k)
{
let start = 0, endval = 0, p = 1,
countOnes = 0, res = 0;
while (endval < n)
{
p *= (arr[endval]);
if (p > k)
{
while (start <= endval && p > k)
{
p = parseInt(p / arr[start], 10);
start++;
}
}
if (p == k)
{
countOnes = 0;
while (endval + 1 < n && arr[endval + 1] == 1)
{
countOnes++;
endval++;
}
res += (countOnes + 1);
while (start <= endval &&
arr[start] == 1 && k != 1)
{
res += (countOnes + 1);
start++;
}
p = parseInt(p / arr[start], 10);
start++;
}
endval++;
}
return res;
}
let arr1 = [ 2, 1, 1, 1, 3, 1, 1, 4 ];
let n1 = arr1.length;
let k = 1;
if (k != 1)
document.write(findSubarrayCount(arr1, n1, k) + "</br>" );
else
document.write(countOne(arr1, n1) + "</br>" );
let arr2 = [ 2, 1, 1, 1, 4, 5 ];
let n2 = arr2.length;
k = 4;
if (k != 1)
document.write(findSubarrayCount(arr2, n2, k) + "</br>" );
else
document.write(countOne(arr2, n2) + "</br>" );
</script>
|
Time Complexity: O(n)
Auxiliary Space : O(1)
Last Updated :
05 Aug, 2022
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