We have an array of integers and a set of range queries. For every query, we need to find product of elements in the given range.
Example:
Input : arr[] = {5, 10, 15, 20, 25} queries[] = {(3, 5), (2, 2), (2, 3)} Output : 7500, 10, 150 7500 = 15 x 20 x 25 10 = 10 150 = 10 x 15
A naive approach would be to navigate through each element in the array from lower to upper and multiply all the elements to find the product. Time complexity would be O(n) per query.
An efficient approach would be to take another array and store the product of all elements upto element with index i in the i’th index of the array. But there is a catch here. If any of the elements in the array is 0, the dp array will contain 0 from then onwards. Hence we should be getting 0 as our answer even if that 0 does not lie in the range between lower and upper.
To solve this case, we have used another array called countZeros which will contain the number of ‘0’s the array contains upto index i.
Now if countZeros[upper]-countZeros[lower] > 0, it signifies that there is a 0 in that range and hence the answer would be 0. Else we would proceed accordingly.
// C++ program for array // range product queries #include<bits/stdc++.h> using namespace std;
// Answers product queries // given as two arrays // lower[] and upper[]. void findProduct( long arr[], int lower[],
int upper[], int n, int n1)
{ long preProd[n];
int countZeros[n];
long prod = 1; // stores the product
// keeps count of zeros
int count = 0;
for ( int i = 0; i < n; i++)
{
// if arr[i] is 0, we increment
// count and do not multiply
// it with the product
if (arr[i] == 0)
count++;
else
prod *= arr[i];
// store the value of prod in dp
preProd[i] = prod;
// store the value of
// count in countZeros
countZeros[i] = count;
}
// We have preprocessed
// the array, let us
// answer queries now.
for ( int i = 0; i < n1; i++)
{
int l = lower[i];
int u = upper[i];
// range starts from
// first element
if (l == 1)
{
// range does not
// contain any zero
if (countZeros[u - 1] == 0)
cout << (preProd[u - 1]) << endl;
else
cout<<0<<endl;
}
else // range starts from
// any other index
{
// no difference in countZeros
// indicates that there are no
// zeros in the range
if (countZeros[u - 1] -
countZeros[l - 2] == 0)
cout << (preProd[u - 1] /
preProd[l - 2]) << endl;
else // zeros are present in the range
cout << 0 << endl;
}
}
} // Driver code int main()
{ long arr[] ={ 0, 2, 3, 4, 5 };
int lower[] = {1, 2};
int upper[] = {3, 2};
findProduct(arr, lower, upper,5,2);
return 0;
} // This code is contributed // by Arnab Kundu |
// Java program for array range product queries class Product {
// Answers product queries given as two arrays
// lower[] and upper[].
static void findProduct( long [] arr, int [] lower,
int [] upper)
{
int n = arr.length;
long [] preProd = new long [n];
int [] countZeros = new int [n];
long prod = 1 ; // stores the product
// keeps count of zeros
int count = 0 ;
for ( int i = 0 ; i < n; i++) {
// if arr[i] is 0, we increment count and
// do not multiply it with the product
if (arr[i] == 0 )
count++;
else
prod *= arr[i];
// store the value of prod in dp
preProd[i] = prod;
// store the value of count in countZeros
countZeros[i] = count;
}
// We have preprocessed the array, let us
// answer queries now.
for ( int i = 0 ; i < lower.length; i++) {
int l = lower[i];
int u = upper[i];
// range starts from first element
if (l == 1 )
{
// range does not contain any zero
if (countZeros[u - 1 ] == 0 )
System.out.println(preProd[u - 1 ]);
else
System.out.println( 0 );
}
else // range starts from any other index
{
// no difference in countZeros indicates that
// there are no zeros in the range
if (countZeros[u - 1 ] - countZeros[l - 2 ] == 0 )
System.out.println(preProd[u - 1 ] / preProd[l - 2 ]);
else // zeros are present in the range
System.out.println( 0 );
}
}
}
public static void main(String[] args)
{
long [] arr = new long [] { 0 , 2 , 3 , 4 , 5 };
int [] lower = { 1 , 2 };
int [] upper = { 3 , 2 };
findProduct(arr, lower, upper);
}
} |
# Python 3 program for array range # product queries # Answers product queries given as # lower[] and upper[]. def findProduct(arr,lower, upper, n, n1):
preProd = [ 0 for i in range (n)]
countZeros = [ 0 for i in range (n)]
prod = 1
# stores the product
# keeps count of zeros
count = 0
for i in range ( 0 , n, 1 ):
# if arr[i] is 0, we increment
# count and do not multiply
# it with the product
if (arr[i] = = 0 ):
count + = 1
else :
prod * = arr[i]
# store the value of prod in dp
preProd[i] = prod
# store the value of count
# in countZeros
countZeros[i] = count
# We have preprocessed the array,
# let us answer queries now.
for i in range ( 0 , n1, 1 ):
l = lower[i]
u = upper[i]
# range starts from first element
if (l = = 1 ):
# range does not contain any zero
if (countZeros[u - 1 ] = = 0 ):
print ( int (preProd[u - 1 ]))
else :
print ( 0 )
else :
# range starts from any other index
# no difference in countZeros indicates
# that there are no zeros in the range
if (countZeros[u - 1 ] -
countZeros[l - 2 ] = = 0 ):
print ( int (preProd[u - 1 ] /
preProd[l - 2 ]))
else :
# zeros are present in the range
print ( 0 )
# Driver code if __name__ = = '__main__' :
arr = [ 0 , 2 , 3 , 4 , 5 ]
lower = [ 1 , 2 ]
upper = [ 3 , 2 ]
findProduct(arr, lower, upper, 5 , 2 )
# This code is contributed by # Sahil_Shelangia |
// C# program for array // range product queries using System;
class GFG
{ // Answers product queries // given as two arrays // lower[] and upper[]. static void findProduct( long [] arr,
int [] lower,
int [] upper)
{ int n = arr.Length;
long [] preProd = new long [n];
int [] countZeros = new int [n];
long prod = 1; // stores the product
// keeps count of zeros
int count = 0;
for ( int i = 0; i < n; i++)
{
// if arr[i] is 0, we increment
// count and do not multiply it
// with the product
if (arr[i] == 0)
count++;
else
prod *= arr[i];
// store the value
// of prod in dp
preProd[i] = prod;
// store the value of
// count in countZeros
countZeros[i] = count;
}
// We have preprocessed
// the array, let us
// answer queries now.
for ( int i = 0;
i < lower.Length; i++)
{
int l = lower[i];
int u = upper[i];
// range starts from
// first element
if (l == 1)
{
// range does not
// contain any zero
if (countZeros[u - 1] == 0)
Console.WriteLine(preProd[u - 1]);
else
Console.WriteLine(0);
}
// range starts from
// any other index
else
{
// no difference in countZeros
// indicates that there are no
// zeros in the range
if (countZeros[u - 1] -
countZeros[l - 2] == 0)
Console.WriteLine(preProd[u - 1] /
preProd[l - 2]);
// zeros are present
// in the range
else
Console.WriteLine(0);
}
}
} // Driver Code public static void Main()
{ long [] arr = {0, 2, 3, 4, 5};
int [] lower = {1, 2};
int [] upper = {3, 2};
findProduct(arr, lower, upper);
} } // This code is contributed // by chandan_jnu. |
<?php // PHP program for array range product queries // Answers product queries given as two arrays // lower[] and upper[]. function findProduct( $arr , $lower , $upper )
{ $n = sizeof( $arr );
$preProd = array ( $n );
$countZeros = array ( $n );
$prod = 1; // stores the product
// keeps count of zeros
$count = 0;
for ( $i = 0; $i < $n ; $i ++)
{
// if arr[i] is 0, we increment count and
// do not multiply it with the product
if ( $arr [ $i ] == 0)
$count ++;
else
$prod *= $arr [ $i ];
// store the value of prod in dp
$preProd [ $i ] = $prod ;
// store the value of count in countZeros
$countZeros [ $i ] = $count ;
}
// We have preprocessed the array, let us
// answer queries now.
for ( $i = 0; $i < sizeof( $lower ); $i ++) {
$l = $lower [ $i ];
$u = $upper [ $i ];
// range starts from first element
if ( $l == 1)
{
// range does not contain any zero
if ( $countZeros [ $u - 1] == 0)
echo ( $preProd [ $u - 1] );
else
echo (0);
}
else // range starts from any other index
{
// no difference in countZeros indicates that
// there are no zeros in the range
if ( $countZeros [ $u - 1] - $countZeros [ $l - 2] == 0)
echo ( $preProd [ $u - 1] / $preProd [ $l - 2]);
else // zeros are present in the range
echo (0);
}
echo "\n" ;
}
} // Driver Code $arr = array (0, 2, 3, 4, 5 );
$lower = array (1, 2);
$upper = array (3, 2);
findProduct( $arr , $lower , $upper );
// This code is contributed // by Mukul Singh |
<script> // JavaScript program for array
// range product queries
// Answers product queries
// given as two arrays
// lower[] and upper[].
function findProduct(arr, lower, upper)
{
let n = arr.length;
let preProd = new Array(n);
preProd.fill(0);
let countZeros = new Array(n);
countZeros.fill(0);
let prod = 1; // stores the product
// keeps count of zeros
let count = 0;
for (let i = 0; i < n; i++)
{
// if arr[i] is 0, we increment
// count and do not multiply it
// with the product
if (arr[i] == 0)
count++;
else
prod *= arr[i];
// store the value
// of prod in dp
preProd[i] = prod;
// store the value of
// count in countZeros
countZeros[i] = count;
}
// We have preprocessed
// the array, let us
// answer queries now.
for (let i = 0;
i < lower.length; i++)
{
let l = lower[i];
let u = upper[i];
// range starts from
// first element
if (l == 1)
{
// range does not
// contain any zero
if (countZeros[u - 1] == 0)
document.write(preProd[u - 1] + "</br>" );
else
document.write(0 + "</br>" );
}
// range starts from
// any other index
else
{
// no difference in countZeros
// indicates that there are no
// zeros in the range
if (countZeros[u - 1] -
countZeros[l - 2] == 0)
document.write(
(preProd[u - 1] / preProd[l - 2]) +
"</br>" );
// zeros are present
// in the range
else
document.write(0 + "</br>" );
}
}
}
let arr = [0, 2, 3, 4, 5];
let lower = [1, 2];
let upper = [3, 2];
findProduct(arr, lower, upper);
</script> |
0 2
Time complexity: O(n) during creation of dp array and O(1) per query
Auxiliary Space: O(n)