Find the number in a range having maximum product of the digits

• Difficulty Level : Medium
• Last Updated : 09 Dec, 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 significant 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 using namespace std; // Returns the product of digits of number xint 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 digitsint 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 Codeint 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 xstatic 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 digitsstatic 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 Codepublic 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 xdef product(x) :         prod = 1    while (x) :        prod *= (x % 10)        x //= 10;         return prod # This function returns the number having# maximum product of the digitsdef 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 Codeif __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// digitsusing System; class GFG{     // Returns the product of digits of number xstatic 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 digitsstatic 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 Codestatic 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

 = \$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?>

Javascript


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 using namespace std;#define int long long int // pair of array to store product and number// dp[pos][tight1][tight2][start]pair dp; 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.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 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 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 maximum product    // initially 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    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 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 approachimport 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 maximum product    // initially 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;         // 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

Python3

 # Python3 program for the above approach  # pair of array to store product and number# dp[pos][tight1][tight2][start]dp=[[[[[0,0] for _ in range(2)] for _ in range(2)]for _ in range(2)]for _ in range(20)] def recur(l, r, pos, ta, tb, st):     # Base case if pos is equal    # to l or r size return    # pair:1,""    if (pos == len(l)) :        return (1, "")          # look up condition    if (dp[pos][ta][tb][st] != -1):        return dp[pos][ta][tb][st]     # Lower bound condition    start = ord(l[pos]) - ord('0') if ta else  0     # Upper bound condition    end = ord(r[pos]) - ord('0') if tb  else 9     # To store the maximum product    # initially its is set to -1    ans = -1     # To store the corresponding    # number as number is large    # so store it as a    s = []     for i in range(start,end+1) :         # Multiply with this val        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        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 * val > ans) :             ans = temp * val             # update only if no leading zeroes            # becoz no use to append the leading zeroes            if (i == 0 and st == 0) :                 s = list(temp)                          else :                 s = list(temp)                 s.append(chr(ord('0') + i))    s=''.join(s)                                # while returning memoize the ans    dp[pos][ta][tb][st] = [ans, s]    return dp[pos][ta][tb][st]  def solve(a, b):     # convert l to sting L and r to R ,    # as eger value should be large    L = list(str(a))     R = str(b)     # to make the size of s    # equal append zeroes in    # front of L    if (len(L) < len(R)) :         L=L[::-1]         while (len(L) < len(R)) :             L.append('0')                  L=L[::-1]    L=''.join(L)     # initialize dp    # as it is pair of array so memset will not work    for i in range(20) :         for j in range(2) :             for k in range(2) :                 for l in range(2):                     dp[i][j][k][l] = -1                                                 # as we have to return pair second value    # it's that number which gaves maximum product    # initially pos=0,ta=1,tb=1,start=0(becoz number is not    # started yet)     ans = recur(L, R, 0, 1, 1, 0)     # reverse it becoz we were appending from right to left    # in recursive call    ans=ans[::-1]     return [ans, ans]  if __name__=='__main__':     # take l and r as input     l = 52; r = 62    print("l=",l)    print("r=",r)    ans = solve(l, r)    print("Maximum Product:",ans)    print("Number which gave maximum product:",ans)
Output
l= 52
r= 62
Maximum Product: 45
Number which gave maximum product: 59

Time Complexity: O(logN), where N is the maximum number between L and R.
Auxiliary Space: O(logN)

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