Count numbers in given range such that sum of even digits is greater than sum of odd digits

Given two integers L and R denoting a range [L, R]. The task is to find the total count of numbers in the given range [L,R] whose sum of even digits is greater than the sum of odd digits.

Examples:

Input : L=2 R=10
Output : 4
Numbers having the property that sum of even
digits is greater than sum of odd digits are: 2, 4, 6, 8

Input : L=2 R=17
Output : 7

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Prerequisites: Digit-DP

Approach:
Firstly, count the required numbers up to R i.e. in the range [0, R]. To reach the answer in the range [L, R] solve for the range from zero to R and then subtracting the answer for the range from zero to L – 1. Define the DP states as follows:

Consider the number as a sequence of digits, one state is the position at which we are currently at. This position can have values from 0 to 18 if we are dealing with the numbers up to 10^18. In each recursive call, try to build the sequence from left to right by placing a digit from 0 to 9.
First state is the sum of the even digits that has been placed so far.
Second state is the sum of the odd digits that has been placed so far.
Another state is the boolean variable tight which tells the number we are trying to build has already become smaller than R so that in the upcoming recursive calls we can place any digit from 0 to 9. If the number has not become smaller, the maximum limit of digit we can place is the digit at the current position in R.

Below is the implementation of the above approach:

C++

// C++ code to count number in the range 
// having the sum of even digits greater 
// than the sum of odd digits 
#include <bits/stdc++.h> 
  
// as the number can be up to 10^18 
#define int long long 
  
using namespace std; 
  
vector<int> v; 
  
int dp[18][180][180][2]; 
  
int memo(int index, int evenSum, 
                      int oddSum, int tight) 
{ 
    // Base Case 
  
    if (index == v.size()) { 
        // check if condition satisfied or not 
        if (evenSum > oddSum) 
            return 1; 
        else
            return 0; 
    } 
  
    // If this result is already computed 
    // simply return it 
    if (dp[index][evenSum][oddSum][tight] != -1) 
        return dp[index][evenSum][oddSum][tight]; 
  
    // Maximum limit upto which we can place 
    // digit. If tight is 0, means number has 
    // already become smaller so we can place 
    // any digit, otherwise num[pos] 
  
    int limit = (tight) ? v[index] : 9; 
  
    int ans = 0; 
  
    for (int d = 0; d <= limit; d++) { 
        int currTight = 0; 
  
        if (d == v[index]) 
            currTight = tight; 
  
        // if current digit is odd 
        if (d % 2 != 0) 
            ans += memo(index + 1, evenSum, 
                        oddSum + d, currTight); 
  
        // if current digit is even 
        else
            ans += memo(index + 1, evenSum + d, 
                        oddSum, currTight); 
    } 
  
    dp[index][evenSum][oddSum][tight] = ans; 
    return ans; 
} 
// Function to convert n into its 
// digit vector and uses memo() function 
// to return the required count 
int CountNum(int n) 
{ 
    v.clear(); 
    while (n) { 
        v.push_back(n % 10); 
        n = n / 10; 
    } 
  
    reverse(v.begin(), v.end()); 
  
    // Initialize DP 
    memset(dp, -1, sizeof(dp)); 
    return memo(0, 0, 0, 1); 
} 
  
// Driver Code 
  
int32_t main() 
{ 
    int L, R; 
    L = 2; 
    R = 10; 
    cout << CountNum(R) - CountNum(L - 1) << "\n"; 
    return 0; 
} 

Java

// Java code to count number in the range 
// having the sum of even digits greater 
// than the sum of odd digits 
import java.util.*; 
  
class GFG 
{ 
  
    static Vector<Integer> v = new Vector<>(); 
  
    static int[][][][] dp = new int[18][180][180][2]; 
  
    static int memo(int index, int evenSum, 
    int oddSum, int tight) 
    { 
        // Base Case 
  
        if (index == v.size())  
        { 
            // check if condition satisfied or not 
            if (evenSum > oddSum)  
            { 
                return 1; 
            }  
            else
            { 
                return 0; 
            } 
        } 
  
        // If this result is already computed 
        // simply return it 
        if (dp[index][evenSum][oddSum][tight] != -1)  
        { 
            return dp[index][evenSum][oddSum][tight]; 
        } 
  
        // Maximum limit upto which we can place 
        // digit. If tight is 0, means number has 
        // already become smaller so we can place 
        // any digit, otherwise num[pos] 
        int limit = (tight > 0) ? v.get(index) : 9; 
  
        int ans = 0; 
  
        for (int d = 0; d <= limit; d++)  
        { 
            int currTight = 0; 
  
            if (d == v.get(index))  
            { 
                currTight = tight; 
            } 
  
            // if current digit is odd 
            if (d % 2 != 0)  
            { 
                ans += memo(index + 1, evenSum, 
                        oddSum + d, currTight); 
            } // if current digit is even 
            else
            { 
                ans += memo(index + 1, evenSum + d, 
                        oddSum, currTight); 
            } 
        } 
  
        dp[index][evenSum][oddSum][tight] = ans; 
        return ans; 
    } 
    // Function to convert n into its 
    // digit vector and uses memo() function 
    // to return the required count 
  
    static int CountNum(int n)  
    { 
        v.clear(); 
        while (n > 0)  
        { 
            v.add(n % 10); 
            n = n / 10; 
        } 
  
        Collections.reverse(v); 
  
        // Initialize DP 
        for (int i = 0; i < 18; i++) 
        { 
            for (int j = 0; j < 180; j++)  
            { 
                for (int k = 0; k < 180; k++)  
                { 
                    for (int l = 0; l < 2; l++) 
                    { 
                        dp[i][j][k][l] = -1; 
                    } 
                } 
            } 
        } 
  
        return memo(0, 0, 0, 1); 
    } 
  
    // Driver Code 
    public static void main(String[] args)  
    { 
        int L, R; 
        L = 2; 
        R = 10; 
        System.out.println(CountNum(R) - CountNum(L - 1)); 
  
    } 
} 
  
// This code is contributed by Princi Singh 

Python3

# Python code to count number in the range 
# having the sum of even digits greater 
# than the sum of odd digits 
  
def memo(index, evenSum, oddSum, tight): 
  
    # Base Case 
    if index == len(v): 
  
        # check if condition satisfied or not 
        if evenSum > oddSum: 
            return 1
        else: 
            return 0
  
    # If this result is already computed 
    # simply return it 
    if dp[index][evenSum][oddSum][tight] != -1: 
        return dp[index][evenSum][oddSum][tight] 
  
    # Maximum limit upto which we can place 
    # digit. If tight is 0, means number has 
    # already become smaller so we can place 
    # any digit, otherwise num[index] 
    limit = v[index] if tight else 9
  
    ans = 0
  
    for d in range(limit + 1): 
        currTight = 0
  
        if d == v[index]: 
            currTight = tight 
  
        # if current digit is odd 
        if d % 2 != 0: 
            ans += memo(index + 1, evenSum,  
                        oddSum + d, currTight) 
  
        # if current digit is even 
        else: 
            ans += memo(index + 1, evenSum + d,  
                            oddSum, currTight) 
  
    dp[index][evenSum][oddSum][tight] = ans 
    return ans 
  
# Function to convert n into its digit vector 
# and uses memo() function to return the 
# required count 
def countNum(n): 
    global dp, v 
  
    v.clear() 
    num = [] 
    while n: 
        v.append(n % 10) 
        n //= 10
  
    v.reverse() 
  
    # Initialize dp 
    dp = [[[[-1, -1] for i in range(180)] for j in range(180)] 
        for k in range(18)] 
    return memo(0, 0, 0, 1) 
  
# Driver Code 
if __name__ == "__main__": 
    dp = [] 
    v = [] 
  
    L = 2
    R = 10
    print(countNum(R) - countNum(L - 1)) 
  
# This code is contributed by 
# sanjeev2552 

C#

// C# code to count number in the range 
// having the sum of even digits greater 
// than the sum of odd digits 
using System.Collections.Generic; 
using System; 
  
class GFG 
{ 
  
    static List<int> v = new List<int>(); 
  
    static int [,,,]dp = new int[18,180,180,2]; 
  
    static int memo(int index, int evenSum, 
    int oddSum, int tight) 
    { 
        // Base Case 
  
        if (index == v.Count)  
        { 
            // check if condition satisfied or not 
            if (evenSum > oddSum)  
            { 
                return 1; 
            }  
            else
            { 
                return 0; 
            } 
        } 
  
        // If this result is already computed 
        // simply return it 
        if (dp[index,evenSum,oddSum,tight] != -1)  
        { 
            return dp[index,evenSum,oddSum,tight]; 
        } 
  
        // Maximum limit upto which we can place 
        // digit. If tight is 0, means number has 
        // already become smaller so we can place 
        // any digit, otherwise num[pos] 
        int limit = (tight > 0) ? v[index] : 9; 
  
        int ans = 0; 
  
        for (int d = 0; d <= limit; d++)  
        { 
            int currTight = 0; 
  
            if (d == v[index])  
            { 
                currTight = tight; 
            } 
  
            // if current digit is odd 
            if (d % 2 != 0)  
            { 
                ans += memo(index + 1, evenSum, 
                        oddSum + d, currTight); 
            } // if current digit is even 
            else
            { 
                ans += memo(index + 1, evenSum + d, 
                        oddSum, currTight); 
            } 
        } 
  
        dp[index,evenSum,oddSum,tight] = ans; 
        return ans; 
    } 
      
    // Function to convert n into its 
    // digit vector and uses memo() function 
    // to return the required count 
    static int CountNum(int n)  
    { 
        v.Clear(); 
        while (n > 0)  
        { 
            v.Add(n % 10); 
            n = n / 10; 
        } 
  
        v.Reverse(); 
  
        // Initialize DP 
        for (int i = 0; i < 18; i++) 
        { 
            for (int j = 0; j < 180; j++)  
            { 
                for (int k = 0; k < 180; k++)  
                { 
                    for (int l = 0; l < 2; l++) 
                    { 
                        dp[i,j,k,l] = -1; 
                    } 
                } 
            } 
        } 
  
        return memo(0, 0, 0, 1); 
    } 
  
    // Driver Code 
    public static void Main(String[] args)  
    { 
        int L, R; 
        L = 2; 
        R = 10; 
        Console.WriteLine(CountNum(R) - CountNum(L - 1)); 
  
    } 
} 
  
/* This code is contributed by PrinciRaj1992 */

Output:

Time Complexity : There would be at max 18*(180)*(180)*2 computations when 0 < a,b < 10¹⁸

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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