{"id":27209,"date":"2018-01-05T20:31:54","date_gmt":"2018-01-05T15:01:54","guid":{"rendered":"https:\/\/www.wikitechy.com\/technology\/?p=27209"},"modified":"2018-01-05T20:31:54","modified_gmt":"2018-01-05T15:01:54","slug":"c-programming-count-of-n-digit-numbers-whose-sum-of-digits-equals-to-given-sum","status":"publish","type":"post","link":"https:\/\/www.wikitechy.com\/technology\/c-programming-count-of-n-digit-numbers-whose-sum-of-digits-equals-to-given-sum\/","title":{"rendered":"Cpp Programming – Count of n digit numbers whose sum of digits equals to given sum"},"content":{"rendered":"

Given two integers \u2018n\u2019 and \u2018sum\u2019, find count of all n digit numbers with sum of digits as \u2018sum\u2019. Leading 0\u2019s are not counted as digits.
\n1 <= n <= 100 and 1 <= sum <= 50000<\/span><\/p>\n

Example:<\/p>\n

Input:  n = 2, sum = 2\r\nOutput: 2\r\nExplanation: Numbers are 11 and 20\r\n\r\nInput:  n = 2, sum = 5\r\nOutput: 5\r\nExplanation: Numbers are 14, 23, 32, 41 and 50\r\n\r\nInput:  n = 3, sum = 6\r\nOutput: 21<\/pre>\n
The idea is simple, we subtract all values from 0 to 9 from given sum and recur for sum minus that digit. Below is recursive formula.<\/p>\n
    countRec(n, sum) = \u2211countRec(n-1, sum-x)\r\n                            where 0 =< x <= 9 and\r\n                                 sum-x >= 0\r\n\r\n    One important observation is, leading 0's must be\r\n    handled explicitly as they are not counted as digits.\r\n    So our final count can be written as below.\r\n    finalCount(n, sum) = \u2211countRec(n-1, sum-x)\r\n                           where 1 =< x <= 9 and\r\n                                 sum-x >= 0<\/pre>\n[ad type=”banner”]\n
Below is a simple recursive solution based on above recursive formula.<\/p>\n
 
 C++<\/span> <\/div> 
\/\/ A recursive program to count numbers with sum\/\/ of digits as given 'sum'#include<bits\/stdc++.h>using namespace std; \/\/ Recursive function to count 'n' digit numbers\/\/ with sum of digits as 'sum'. This function\/\/ considers leading 0's also as digits, that is\/\/ why not directly calledunsigned long long int countRec(int n, int sum){    \/\/ Base case    if (n == 0)       return sum == 0;     \/\/ Initialize answer    unsigned long long int ans = 0;     \/\/ Traverse through every digit and count    \/\/ numbers beginning with it using recursion    for (int i=0; i<=9; i++)       if (sum-i >= 0)          ans += countRec(n-1, sum-i);     return 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, int sum){    \/\/ Initialize final answer    unsigned long long int ans = 0;     \/\/ Traverse through every digit from 1 to    \/\/ 9 and count numbers beginning with it    for (int i = 1; i <= 9; i++)      if (sum-i >= 0)         ans += countRec(n-1, sum-i);     return ans;} \/\/ Driver programint main(){    int n = 2, sum = 5;    cout << finalCount(n, sum);    return 0;}<\/code><\/pre> <\/div>\n
Output:<\/p>\n
5<\/pre>\n
The time complexity of above solution is exponential. If we draw the complete recursion tree, we can observer that many subproblems are solved again and again. For example, if we start with n = 3 and sum = 10, we can reach n = 1, sum = 8, by considering digit sequences 1,1 or 2, 0.
\nSince same suproblems are called again, this problem has Overlapping Subprolems property. So min square sum problem has both properties of a dynamic programming problem.<\/p>\n
Below is Memoization based the implementation.<\/p>\n
 
 C++<\/span> <\/div> 
\/\/ A memoization based recursive program to count \/\/ numbers with sum of n as given 'sum'#include<bits\/stdc++.h>using namespace std; \/\/ A lookup table used for memoizationunsigned long long int lookup[101][50001]; \/\/ Memoizatiob based implementation of recursive\/\/ functionunsigned long long int countRec(int n, int sum){    \/\/ Base case    if (n == 0)       return sum == 0;     \/\/ If this subproblem is already evaluated,    \/\/ return the evaluated value    if (lookup[n][sum] != -1)       return lookup[n][sum];     \/\/ Initialize answer    unsigned long long int ans = 0;     \/\/ Traverse through every digit and    \/\/ recursively count numbers beginning    \/\/ with it    for (int i=0; i<10; i++)       if (sum-i >= 0)          ans += countRec(n-1, sum-i);     return lookup[n][sum] = ans;} \/\/ This is mainly a wrapper over countRec. It\/\/ explicitly handles leading digit and calls\/\/ countRec() for remaining n.unsigned long long int finalCount(int n, int sum){    \/\/ Initialize all entries of lookup table    memset(lookup, -1, sizeof lookup);     \/\/ Initialize final answer    unsigned long long int ans = 0;     \/\/ Traverse through every digit from 1 to    \/\/ 9 and count numbers beginning with it    for (int i = 1; i <= 9; i++)      if (sum-i >= 0)         ans += countRec(n-1, sum-i);    return ans;} \/\/ Driver programint main(){    int n = 3, sum = 5;    cout << finalCount(n, sum);    return 0;}<\/code><\/pre> <\/div>\n[ad type=”banner”]\n
Output:<\/p>\n
5<\/pre>\n","protected":false},"excerpt":{"rendered":"
C++ Programming – Count of n digit numbers whose sum of digits equals to given sum – Dynamic Programming Given two integers \u2018n\u2019 and \u2018sum\u2019, find count of all<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[83515,1,70145],"tags":[81124,72844,72850,72839,81132,81128,81119,72852,81131,72855,72853],"class_list":["post-27209","post","type-post","status-publish","format-standard","hentry","category-c-programming-3","category-coding","category-dynamic-programming","tag-a-problem-from-a-programming-competition","tag-dynamic-programming-software","tag-explain-dynamic-programming","tag-how-to-solve-dynamic-programming-problems","tag-number-of-occurrences-of-the-digit-1-in-the-numbers-from-0-to-n","tag-numbers-of-length-n-and-value-less-than-k","tag-print-all-n-digit-numbers-whose-sum-of-digits-equals-to-given-sum","tag-problems-on-dynamic-programming","tag-program-to-find-sum-of-digits-of-a-number-in-c","tag-simple-dynamic-programming-example","tag-types-of-dynamic-programming"],"_links":{"self":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/27209","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/comments?post=27209"}],"version-history":[{"count":0,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/27209\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/media?parent=27209"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/categories?post=27209"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/tags?post=27209"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}