Given a number n, we need to count the total number of n digit numbers such that the sum of even digits is 1 more than the sum of odd digits. Here even and odd means positions of digits are like array indexes, for example, the leftmost (or leading) digit is considered as even digit, next to leftmost is considered as odd, and so on.
Example:
Input: n = 2 Output: Required Count of 2 digit numbers is 9 Explanation : 10, 21, 32, 43, 54, 65, 76, 87, 98. Input: n = 3 Output: Required Count of 3 digit numbers is 54 Explanation: 100, 111, 122, ......, 980
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This problem is mainly an extension of Count of n digit numbers whose sum of digits equals to given sum. Here the solution of subproblems depends on four variables: digits, esum (current even sum), osum (current odd sum), isEven(A flag to indicate whether the current digit is even or odd).
Below is Memoization based solution for the same.
// A memoization based recursive program to count numbers // with difference between odd and even digit sums as 1 #include<bits/stdc++.h> using namespace std;
// A lookup table used for memoization. unsigned long long int lookup[50][1000][1000][2];
// Memoization based recursive function to count numbers // with even and odd digit sum difference as 1. This function // considers leading zero as a digit unsigned long long int countRec( int digits, int esum,
int osum, bool isOdd, int n)
{ // Base Case
if (digits == n)
return (esum - osum == 1);
// If current subproblem is already computed
if (lookup[digits][esum][osum][isOdd] != -1)
return lookup[digits][esum][osum][isOdd];
// Initialize result
unsigned long long int ans = 0;
// If the current digit is odd, then add it to odd sum and recur
if (isOdd)
for ( int i = 0; i <= 9; i++)
ans += countRec(digits+1, esum, osum+i, false , n);
else // Add to even sum and recur
for ( int i = 0; i <= 9; i++)
ans += countRec(digits+1, esum+i, osum, true , n);
// Store current result in lookup table and return the same
return lookup[digits][esum][osum][isOdd] = ans;
} // This is mainly a wrapper over countRec. It // explicitly handles leading digit and calls // countRec() for remaining digits. unsigned long long int finalCount( int n)
{ // Initialize number digits considered so far
int digits = 0;
// Initialize all entries of lookup table
memset (lookup, -1, sizeof lookup);
// Initialize final answer
unsigned long long int ans = 0;
// Initialize even and odd sums
int esum = 0, osum = 0;
// Explicitly handle first digit and call recursive function
// countRec for remaining digits. Note that the first digit
// is considered as even digit.
for ( int i = 1; i <= 9; i++)
ans += countRec(digits+1, esum + i, osum, true , n);
return ans;
} // Driver program int main()
{ int n = 3;
cout << "Count of " <<n << " digit numbers is " << finalCount(n);
return 0;
} |
// A memoization based recursive // program to count numbers with // difference between odd and // even digit sums as 1 class GFG
{ // A lookup table used for memoization. static int [][][][]lookup = new int [ 50 ][ 1000 ][ 1000 ][ 2 ];
// Memoization based recursive // function to count numbers // with even and odd digit sum // difference as 1. This function // considers leading zero as a digit static int countRec( int digits, int esum,
int osum, int isOdd, int n)
{ // Base Case
if (digits == n)
return (esum - osum == 1 )? 1 : 0 ;
// If current subproblem is already computed
if (lookup[digits][esum][osum][isOdd] != - 1 )
return lookup[digits][esum][osum][isOdd];
// Initialize result
int ans = 0 ;
// If current digit is odd, then
// add it to odd sum and recur
if (isOdd== 1 )
for ( int i = 0 ; i <= 9 ; i++)
ans += countRec(digits+ 1 , esum, osum+i, 0 , n);
else // Add to even sum and recur
for ( int i = 0 ; i <= 9 ; i++)
ans += countRec(digits+ 1 , esum+i, osum, 1 , n);
// Store current result in lookup
// table and return the same
return lookup[digits][esum][osum][isOdd] = ans;
} // This is mainly a wrapper over countRec. It // explicitly handles leading digit and calls // countRec() for remaining digits. static int finalCount( int n)
{ // Initialize number digits considered so far
int digits = 0 ;
// Initialize all entries of lookup table
for ( int i = 0 ; i < 50 ; i++)
for ( int j = 0 ; j < 1000 ; j++)
for ( int k = 0 ; k < 1000 ; k++)
for ( int l = 0 ; l < 2 ; l++)
lookup[i][j][k][l] = - 1 ;
// Initialize final answer
int ans = 0 ;
// Initialize even and odd sums
int esum = 0 , osum = 0 ;
// Explicitly handle first digit and
// call recursive function countRec
// for remaining digits. Note that
// the first digit is considered
// as even digit.
for ( int i = 1 ; i <= 9 ; i++)
ans += countRec(digits+ 1 , esum + i, osum, 1 , n);
return ans;
} // Driver program public static void main(String[] args)
{ int n = 3 ;
System.out.println( "Count of " + n +
" digit numbers is " + finalCount(n));
} } // This code has been contributed by 29AjayKumar |
# A memoization based recursive program to count numbers # with difference between odd and even digit sums as 1 # Memoization based recursive function to count numbers # with even and odd digit sum difference as 1. This function # considers leading zero as a digit def countRec(digits, esum, osum, isOdd, n):
# Base Case
if digits = = n:
return (esum - osum = = 1 )
# If current subproblem is already computed
if lookup[digits][esum][osum][isOdd] ! = - 1 :
return lookup[digits][esum][osum][isOdd]
# Initialize result
ans = 0
# If the current digit is odd,
# then add it to odd sum and recur
if isOdd:
for i in range ( 10 ):
ans + = countRec(digits + 1 , esum,
osum + i, False , n)
# Add to even sum and recur
else :
for i in range ( 10 ):
ans + = countRec(digits + 1 , esum + i,
osum, True , n)
# Store current result in lookup table
# and return the same
lookup[digits][esum][osum][isOdd] = ans
return ans
# This is mainly a wrapper over countRec. It # explicitly handles leading digit and calls # countRec() for remaining digits. def finalCount(n):
global lookup
# Initialize number digits considered so far
digits = 0
# Initialize all entries of lookup table
lookup = [[[[ - 1 , - 1 ] for i in range ( 500 )]
for j in range ( 500 )]
for k in range ( 50 )]
# Initialize final answer
ans = 0
# Initialize even and odd sums
esum = 0
osum = 0
# Explicitly handle first digit and call
# recursive function countRec for remaining digits.
# Note that the first digit is considered as even digit
for i in range ( 1 , 10 ):
ans + = countRec(digits + 1 , esum + i,
osum, True , n)
return ans
# Driver Code if __name__ = = "__main__" :
# A lookup table used for memoization.
lookup = []
n = 3
print ( "Count of %d digit numbers is %d" % (n, finalCount(n)))
# This code is contributed by # sanjeev2552 |
// A memoization based recursive // program to count numbers with // difference between odd and // even digit sums as 1 using System;
class GFG
{ // A lookup table used for memoization. static int [,,,]lookup = new int [50,1000,1000,2];
// Memoization based recursive // function to count numbers // with even and odd digit sum // difference as 1. This function // considers leading zero as a digit static int countRec( int digits, int esum,
int osum, int isOdd, int n)
{ // Base Case
if (digits == n)
return (esum - osum == 1)?1:0;
// If current subproblem is already computed
if (lookup[digits,esum,osum,isOdd] != -1)
return lookup[digits,esum,osum,isOdd];
// Initialize result
int ans = 0;
// If current digit is odd, then
// add it to odd sum and recur
if (isOdd==1)
for ( int i = 0; i <= 9; i++)
ans += countRec(digits+1, esum, osum+i, 0, n);
else // Add to even sum and recur
for ( int i = 0; i <= 9; i++)
ans += countRec(digits+1, esum+i, osum, 1, n);
// Store current result in lookup
// table and return the same
return lookup[digits,esum,osum,isOdd] = ans;
} // This is mainly a wrapper over countRec. It // explicitly handles leading digit and calls // countRec() for remaining digits. static int finalCount( int n)
{ // Initialize number digits considered so far
int digits = 0;
// Initialize all entries of lookup table
for ( int i = 0; i < 50; i++)
for ( int j = 0; j < 1000; j++)
for ( int k = 0; k < 1000; k++)
for ( int l = 0; l < 2; l++)
lookup[i,j,k,l] = -1;
// Initialize final answer
int ans = 0;
// Initialize even and odd sums
int esum = 0, osum = 0;
// Explicitly handle first digit and
// call recursive function countRec
// for remaining digits. Note that
// the first digit is considered
// as even digit.
for ( int i = 1; i <= 9; i++)
ans += countRec(digits+1, esum + i, osum, 1, n);
return ans;
} // Driver code public static void Main(String[] args)
{ int n = 3;
Console.WriteLine( "Count of " + n +
" digit numbers is " + finalCount(n));
} } // This code is contributed by Rajput-Ji |
<script> // A memoization based recursive // program to count numbers with // difference between odd and // even digit sums as 1 // A lookup table used for memoization. let lookup = new Array(50);
for (let i=0;i<50;i++)
{ lookup[i]= new Array(1000);
for (let j=0;j<1000;j++)
{
lookup[i][j]= new Array(1000);
for (let k=0;k<1000;k++)
{
lookup[i][j][k]= new Array(2);
}
}
} // Memoization based recursive // function to count numbers // with even and odd digit sum // difference as 1. This function // considers leading zero as a digit function countRec(digits,esum,osum,isOdd,n)
{ // Base Case
if (digits == n)
return (esum - osum == 1)?1:0;
// If current subproblem is already computed
if (lookup[digits][esum][osum][isOdd] != -1)
return lookup[digits][esum][osum][isOdd];
// Initialize result
let ans = 0;
// If current digit is odd, then
// add it to odd sum and recur
if (isOdd==1)
for (let i = 0; i <= 9; i++)
ans += countRec(digits+1, esum, osum+i, 0, n);
else // Add to even sum and recur
for (let i = 0; i <= 9; i++)
ans += countRec(digits+1, esum+i, osum, 1, n);
// Store current result in lookup
// table and return the same
return lookup[digits][esum][osum][isOdd] = ans;
} // This is mainly a wrapper over countRec. It // explicitly handles leading digit and calls // countRec() for remaining digits. function finalCount(n)
{ // Initialize number digits considered so far
let digits = 0;
// Initialize all entries of lookup table
for (let i = 0; i < 50; i++)
for (let j = 0; j < 1000; j++)
for (let k = 0; k < 1000; k++)
for (let l = 0; l < 2; l++)
lookup[i][j][k][l] = -1;
// Initialize final answer
let ans = 0;
// Initialize even and odd sums
let esum = 0, osum = 0;
// Explicitly handle first digit and
// call recursive function countRec
// for remaining digits. Note that
// the first digit is considered
// as even digit.
for (let i = 1; i <= 9; i++)
ans += countRec(digits+1, esum + i, osum, 1, n);
return ans;
} // Driver program let n = 3; document.write( "Count of " + n +
" digit numbers is " + finalCount(n));
// This code is contributed by rag2127 </script> |
Output:
Count of 3 digit numbers is 54
Thanks to Gaurav Ahirwar for providing above solution.