Count total number of N digit numbers such that the difference between sum of even and odd digits is 1
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.
C++
// 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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
Javascript
<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.
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