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Find the number in a range having maximum product of the digits
  • Difficulty Level : Medium
  • Last Updated : 23 Apr, 2021

Given a range represented by two positive integers L and R. Find the number lying in the range having the maximum product of the digits.
Examples: 

Input : L = 1, R = 10
Output : 9

Input : L = 51, R = 62
Output : 59

Approach : The key idea here is to iterate over the digits of the number R starting from the most significant digit. Going from left to right, i.e. from most sigificant digit to the least significant digit, replace the current digit with one less than current digit and replace all the digits after current digit in the number with 9, since the number has already become smaller than R at the current position so we can safely put any number in the following digits to maximize the product of digits. Also, check if the resulting number is greater than L to remain in the range and update the maximum product.
Below is the implementation of the above approach: 

C++




// CPP Program to find the number in a
// range having maximum product of the
// digits
 
#include <bits/stdc++.h>
using namespace std;
 
// Returns the product of digits of number x
int product(int x)
{
    int prod = 1;
    while (x) {
        prod *= (x % 10);
        x /= 10;
    }
    return prod;
}
 
// This function returns the number having
// maximum product of the digits
int findNumber(int l, int r)
{
    // Converting both integers to strings
    string a = to_string(l);
    string b = to_string(r);
 
    // Let the current answer be r
    int ans = r;
    for (int i = 0; i < b.size(); i++) {
        if (b[i] == '0')
            continue;
 
        // Stores the current number having
        // current digit one less than current
        // digit in b
        string curr = b;
        curr[i] = ((curr[i] - '0') - 1) + '0';
 
        // Replace all following digits with 9
        // to maximise the product
        for (int j = i + 1; j < curr.size(); j++)
            curr[j] = '9';
 
        // Convert string to number
        int num = 0;
        for (auto c : curr)
            num = num * 10 + (c - '0');
 
        // Check if it lies in range and its product
        // is greater than max product
        if (num >= l && product(ans) < product(num))
            ans = num;
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    int l = 1, r = 10;
    cout << findNumber(l, r) << endl;
 
    l = 51, r = 62;
    cout << findNumber(l, r) << endl;
 
    return 0;
}

Java




// Java Program to find the number in a
// range having maximum product of the
// digits
 
class GFG
{
     
// Returns the product of digits of number x
static int product(int x)
{
    int prod = 1;
    while (x > 0)
    {
        prod *= (x % 10);
        x /= 10;
    }
    return prod;
}
 
// This function returns the number having
// maximum product of the digits
static int findNumber(int l, int r)
{
    // Converting both integers to strings
    //string a = l.ToString();
    String b = Integer.toString(r);
 
    // Let the current answer be r
    int ans = r;
    for (int i = 0; i < b.length(); i++)
    {
        if (b.charAt(i) == '0')
            continue;
 
        // Stores the current number having
        // current digit one less than current
        // digit in b
        char[] curr = b.toCharArray();
        curr[i] = (char)(((int)(curr[i] -
                    (int)'0') - 1) + (int)('0'));
 
        // Replace all following digits with 9
        // to maximise the product
        for (int j = i + 1; j < curr.length; j++)
            curr[j] = '9';
 
        // Convert string to number
        int num = 0;
        for (int j = 0; j < curr.length; j++)
            num = num * 10 + (curr[j] - '0');
 
        // Check if it lies in range and its product
        // is greater than max product
        if (num >= l && product(ans) < product(num))
            ans = num;
    }
 
    return ans;
}
 
// Driver Code
public static void main (String[] args)
{
    int l = 1, r = 10;
    System.out.println(findNumber(l, r));
 
    l = 51;
    r = 62;
    System.out.println(findNumber(l, r));
}
}
 
// This code is contributed by chandan_jnu

Python3




# Python3 Program to find the number
# in a range having maximum product
# of the digits
 
# Returns the product of digits
# of number x
def product(x) :
     
    prod = 1
    while (x) :
        prod *= (x % 10)
        x //= 10;
     
    return prod
 
# This function returns the number having
# maximum product of the digits
def findNumber(l, r) :
     
    # Converting both integers to strings
    a = str(l);
    b = str(r);
 
    # Let the current answer be r
    ans = r
     
    for i in range(len(b)) :
        if (b[i] == '0') :
            continue
 
        # Stores the current number having
        # current digit one less than current
        # digit in b
        curr = list(b)
        curr[i] = str(((ord(curr[i]) -
                        ord('0')) - 1) + ord('0'))
 
        # Replace all following digits with 9
        # to maximise the product
        for j in range(i + 1, len(curr)) :
            curr[j] = str(ord('9'))
             
        # Convert string to number
        num = 0
        for c in curr :
            num = num * 10 + (int(c) - ord('0'))
 
        # Check if it lies in range and its
        # product is greater than max product
        if (num >= l and product(ans) < product(num)) :
            ans = num
 
    return ans
 
# Driver Code
if __name__ == "__main__" :
     
    l, r = 1, 10
    print(findNumber(l, r))
 
    l, r = 51, 62
    print(findNumber(l, r))
 
# This code is contributed by Ryuga

C#




// C# Program to find the number in a
// range having maximum product of the
// digits
using System;
 
class GFG
{
     
// Returns the product of digits of number x
static int product(int x)
{
    int prod = 1;
    while (x > 0)
    {
        prod *= (x % 10);
        x /= 10;
    }
    return prod;
}
 
// This function returns the number having
// maximum product of the digits
static int findNumber(int l, int r)
{
    // Converting both integers to strings
    //string a = l.ToString();
    string b = r.ToString();
 
    // Let the current answer be r
    int ans = r;
    for (int i = 0; i < b.Length; i++)
    {
        if (b[i] == '0')
            continue;
 
        // Stores the current number having
        // current digit one less than current
        // digit in b
        char[] curr = b.ToCharArray();
        curr[i] = (char)(((int)(curr[i] -
                    (int)'0') - 1) + (int)('0'));
 
        // Replace all following digits with 9
        // to maximise the product
        for (int j = i + 1; j < curr.Length; j++)
            curr[j] = '9';
 
        // Convert string to number
        int num = 0;
        for (int j = 0; j < curr.Length; j++)
            num = num * 10 + (curr[j] - '0');
 
        // Check if it lies in range and its product
        // is greater than max product
        if (num >= l && product(ans) < product(num))
            ans = num;
    }
 
    return ans;
}
 
// Driver Code
static void Main()
{
    int l = 1, r = 10;
    Console.WriteLine(findNumber(l, r));
 
    l = 51;
    r = 62;
    Console.WriteLine(findNumber(l, r));
}
}
 
// This code is contributed by chandan_jnu

PHP




<?php
// PHP Program to find the number
// in a range having maximum product
// of the digits
 
// Returns the product of digits
// of number x
function product($x)
{
    $prod = 1;
    while ($x)
    {
        $prod *= ($x % 10);
        $x = (int)($x / 10);
    }
     
    return $prod;
}
 
// This function returns the number
// having maximum product of the digits
function findNumber($l, $r)
{
    // Let the current answer be r
    $ans = $r;
     
    // Converting both integers
    // to strings
    $a = strval($l);
    $b = strval($r);
 
    for ($i = 0; $i < strlen($b); $i++)
    {
        if ($b[$i] == '0')
            continue;
 
        // Stores the current number having
        // current digit one less than
        // current digit in b
        $curr = $b;
        $curr[$i] = chr(((ord($curr[$i]) -
                          ord('0')) - 1) +
                          ord('0'));
 
        // Replace all following digits
        // with 9 to maximise the product
        for ($j = $i + 1; $j < strlen($curr); $j++)
            $curr[$j] = '9';
             
        // Convert string to number
        $num = 0;
        for ($c = 0; $c < strlen($curr); $c++)
            $num = $num * 10 + (ord($curr[$c]) -
                                ord('0'));
 
        // Check if it lies in range and its
        // product is greater than max product
        if ($num >= $l and
            product($ans) < product($num))
            $ans = $num;
    }
 
    return $ans;
}
 
// Driver Code
$l = 1;
$r = 10;
print(findNumber($l, $r) . "\n");
 
$l = 51;
$r = 62;
print(findNumber($l, $r));
 
// This code is contributed
// by chandan_jnu
?>
Output: 
9
59

 

Time Complexity: O(18 * 18), if we are dealing with the numbers upto 1018.

Another Approach:  It can be solved using Digit Dp



Key Points of Observation:-

1.  As we know we use tight in digit dp to check whether the range for this digit is restricted or not,same here we will use tight ta and tight tb (basically two tight conditions) ,where ta will tell us

the lower_bound of the digit and tb will tell us the upper_bound of the digit and reason to use two tight values is that we have to calculate the maximum product,it may be the case as:-

max(l,r) ≠ max(r) – max(l-1) and our integer should lie in a range from l to r.

2.  Let suppose the range values as, l=5 and r=15 , so to make size equal we should append the zeroes in front of number after converting to string and taking care of leading zeroes while calculating the answer,

Dp states include:-

1) pos  

  • it will tell the position of index from left in the integer

2) ta    



  • it represents the lower_bound of a digit,we have to make sure number should be greater than or equal to { l }
  • Let suppose we are building a number greater than equal to 0055 and we have created a sequence like 005, so at the 4th place, we can’t put digit less than 5,that will only have digits  between 5 to 9. So for checking this bound, we need ta.
Example : Consider the value of  l = 005
Index   : 0 1 2
digits  : 0 0 5
valid numbers like: 005,006,007,008...
invalid numbers like: ...001,002,003,004

3) tb    

  • the upper_bound of a digit,we have to make sure number should be lesser than or equal to { r }
  • Again let suppose we are building a number lesser than equal to 526 and we have created a sequence like 52, so at the 3rd place, we can’t put digit greater than 6,there we can only place between 0 to 6. So for checking this bound, we need tb
Example : Consider the value of  r = 150
Index   : 0 1 2
digits  : 1 5 0
valid numbers like: ...148,149,150
invalid numbers like: 151,152,153...

4) st      

  • used to check for leading zeroes( as 005 ~ 5)

3.  Constraints: l and r (1 ≤ l ≤ r ≤ 10^18)

Algorithm:

  • We will traverse i from start to end on the basis of tight ta and tight tb as:
start = ta == 1 ? l[ pos ] - '0' : 0;
end = tb ==1 ? r[ pos ] -'0'  : 9;
  • Firstly we will check for leading zeroes as :
if ( st == 0 and i = 0) then multiply with 1,else multiply with i
  • For every position we will calculate the product of sequence and check whether it is the maximum product or not and store the corresponding number
int ans = 0;
for(int i = start; i <= end; i++){
  int val = i;
  if (st==0 and i==0) val = 1;
  ans = max (ans, val * solve (pos+1, ta&(i==start),tb&(i==end) ,st|i>0);
}

C++ implementation:

C++




// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define int long long int
 
// pair of array to store product and number
// dp[pos][tight1][tight2][start]
pair<int, string> dp[20][2][2][2];
 
pair<int, string> recur(string l, string r, int pos, int ta,
                        int tb, int st)
{
 
    // Base case if pos is equal
    // to l or r size return
    // pair{1,""}
    if (pos == l.size()) {
        return { 1, "" };
    }
 
    // look up condition
    if (dp[pos][ta][tb][st].first != -1)
        return dp[pos][ta][tb][st];
 
    // Lower bound condition
    int start = ta ? l[pos] - '0' : 0;
 
    // Upper bound condition
    int end = tb ? r[pos] - '0' : 9;
 
    // To store the maximum product
    // initially its is set to -1
    int ans = -1;
 
    // To store the corresponding
    // number as number is large
    // so store it as a string
    string s = "";
 
    for (int i = start; i <= end; i++) {
 
        // Multiply with this val
        int val = i;
 
        // check for leading zeroes as 00005
        if (st == 0 and i == 0) {
 
            val = 1;
        }
 
        // Recursive call for next
        // position and store it in
        // a pair pair first gives
        // maximum product pair
        // second gives number which
        // gave maximum product
        pair<int, string> temp
 
            = recur(l, r, pos + 1, ta & (i == start),
 
                    tb & (i == end), st | i > 0);
 
        // check if calculated product is greater than
        // previous calculated ans
        if (temp.first * val > ans) {
 
            ans = temp.first * val;
 
            // update string only if no leading zeroes
            // becoz no use to append the leading zeroes
            if (i == 0 and st == 0) {
 
                s = temp.second;
            }
 
            else {
 
                s = temp.second;
 
                s.push_back('0' + i);
            }
        }
    }
 
    // while returning memoize the ans
    return dp[pos][ta][tb][st] = { ans, s };
}
 
pair<int, string> solve(int a, int b)
 
{
 
    // convert int l to sting L and int r to string R ,
    // as integer value should be large
    string L = to_string(a);
 
    string R = to_string(b);
 
    // to make the size of strings
    // equal append zeroes in
    // front of string L
    if (L.size() < R.size()) {
 
        reverse(L.begin(), L.end());
 
        while (L.size() < R.size()) {
 
            L.push_back('0');
        }
 
        reverse(L.begin(), L.end());
    }
 
    // initialize dp
    // as it is pair of array so memset will not work
    for (int i = 0; i < 20; i++) {
 
        for (int j = 0; j < 2; j++) {
 
            for (int k = 0; k < 2; k++) {
 
                for (int l = 0; l < 2; l++) {
 
                    dp[i][j][k][l].first = -1;
                }
            }
        }
    }
 
    // as we have to return pair second value
    // it's that number which gaves mximum product
    // initally pos=0,ta=1,tb=1,start=0(becoz number is not
    // started yet)
 
    pair<int, string> ans = recur(L, R, 0, 1, 1, 0);
 
    // reverse it becoz we were appending from right to left
    // in recursive call
    reverse(ans.second.begin(), ans.second.end());
 
    return { ans.first, ans.second };
}
 
signed main()
 
{
 
    // take l and r as input
 
    int l = 52, r = 62;
    cout << "l= " << l << "\n";
    cout << "r= " << r << "\n";
    pair<int, string> ans = solve(l, r);
    cout << "Maximum Product: " << ans.first << "\n";
    cout << "Number which gave maximum product: "
         << ans.second;
 
    return 0;
}

Java




// JAVA program for the above approach
import java.util.*;
import java.io.*;
import java.math.*;
class GFG
{
   
// pair of array to store product and number
// dp[pos][tight1][tight2][start]
static class pair
 {
    int first;
    String second;
    pair(int first,String second)
      {
         this.first=first;
         this.second=second;
      }
 }
  
static pair dp[][][][];
 
static pair recur(String l, String r, int pos, int ta,
                        int tb, int st)
{
 
    // Base case if pos is equal
    // to l or r size return
    // pair{1,""}
    if (pos == l.length()) {
        return new pair(1,"");
    }
 
    // look up condition
    if (dp[pos][ta][tb][st].first != -1)
        return dp[pos][ta][tb][st];
 
    // Lower bound condition
    int start = ta ==1 ? l.charAt(pos) - '0' : 0;
 
    // Upper bound condition
    int end = tb ==1 ? r.charAt(pos)  - '0' : 9;
 
    // To store the maximum product
    // initially its is set to -1
    int ans = -1;
 
    // To store the corresponding
    // number as number is large
    // so store it as a string
    String s = "";
 
    for (int i = start; i <= end; i++) {
 
        // Multiply with this val
        int val = i;
 
        // check for leading zeroes as 00005
        if (st == 0 && i == 0) {
 
            val = 1;
        }
 
        // Recursive call for next
        // position and store it in
        // a pair pair first gives
        // maximum product pair
        // second gives number which
        // gave maximum product
        pair temp
 
            = recur(l, r, pos + 1, ta==1 ?  (i == start ? 1 : 0) : 0,
 
                    tb==1  ? (i == end ? 1 : 0) : 0, (st | i) > 0 ? 1 : 0);
 
        // check if calculated product is greater than
        // previous calculated ans
        if (temp.first * val > ans) {
 
            ans = temp.first * val;
 
            // update string only if no leading zeroes
            // becoz no use to append the leading zeroes
            if (i == 0 && st == 0) {
 
                s = temp.second;
            }
 
            else {
 
                s = temp.second;
 
                s+=(i);
            }
        }
    }
 
    // while returning memoize the ans
    return dp[pos][ta][tb][st] = new pair(ans, s );
}
static String reverse(String x)
 {
    StringBuilder sb=new StringBuilder("");
    sb.append(x);
    sb.reverse();
    return sb.toString();
 }
static pair solve(int a, int b)
 
{
 
    // convert int l to sting L and int r to string R ,
    // as integer value should be large
    String L = Integer.toString(a);
    String R = Integer.toString(b);
 
 
    // to make the size of strings
    // equal append zeroes in
    // front of string L
    if (L.length() < R.length()) {
 
        L=reverse(L);
 
        while (L.length() < R.length()) {
 
            L += "0";
        }
 
        L=reverse(L);
    }
 
    // initialize dp
    // as it is pair of array so memset will not work
    for (int i = 0; i < 20; i++) {
 
        for (int j = 0; j < 2; j++) {
 
            for (int k = 0; k < 2; k++) {
 
                for (int l = 0; l < 2; l++) {
 
                    dp[i][j][k][l] = new pair(-1,"");
                }
            }
        }
    }
 
    // as we have to return pair second value
    // it's that number which gaves mximum product
    // initally pos=0,ta=1,tb=1,start=0(becoz number is not
    // started yet)
 
    pair ans = recur(L, R, 0, 1, 1, 0);
 
    // reverse it becoz we were appending from right to left
    // in recursive call
    ans.second = reverse(ans.second);
   pair result = new pair(ans.first, ans.second);
    return result;
}
 
public static void main(String args[])
{
 
    // take l and r as input
    int l = 52, r = 62;
    System.out.println("l= "+l );
    System.out.println("r= "+r );
     
    // creation of dp table
    dp = new pair[20][2][2][2];
     
    // call function
    pair ans = solve(l, r);
    System.out.println("Maximum Product: "+ans.first);
    System.out.println("Number which gave maximum product: "+ans.second);
}
}
 
// This code is contributed by Debojyoti Mandal
Output
l= 52
r= 62
Maximum Product: 45
Number which gave maximum product: 59

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