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# Range product queries in an array

• Difficulty Level : Easy
• Last Updated : 28 May, 2021

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++

 `// C++ program for array``// range product queries``#include``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<

## Java

 `// 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);``    ``}``}`

## Python3

 `# 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#

 `// 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

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## Javascript

 ``
Output:
```0
2```

Time complexity: O(n) during creation of dp array and O(1) per query

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