# Minimum count of elements that sums to a given number

Last Updated : 10 Mar, 2022

Given infinite number of elements of form and ( n >= 0 ). The task is to find the minimum count of elements chosen such that there sum is equal to K
Examples:

Input : K = 48
Output :
elements chosen are: (1 + 1 + 1 + 10 + 10 + 25)
Input : 69
Output :
elements chosen are: (1 + 1 + 1 + 1 + 10 + 10 + 10 + 10 + 25)

Recommended Practice

Approach:
There are infinite number of the following elements :
1, 10, 25, 100, 1000, 2500, 10000, 100000, 250000 … and so on.
Greedy Approach won’t work here. For K = 66, by Greedy Approach minimum count will be 9 and chosen elements are 25 + 25 + 10 + 1 + 1 + 1 + 1 + 1 + 1 = 66. But its optimum answer is 6 when these elements are chosen: 25 + 10 + 10 + 10 + 10 + 1 = 66. So, Dynamic Programming will work here. But simple DP cannot be applied because K can go upto 10^9 .
Dynamic Programming approach:

• Precompute the minimum no. of elements chosen that constitutes sum upto 99 and store it in memo array.
• Also, sums upto 99 can only be formed by the combinations of 1, 10 and 25.
• Starting from end of K, iterate over each last 2 digits to find minimum no. of elements chosen that will sum to last two digits.
• Sum them all to find the minimum count.

Illustration of the above approach:
Let’s take K = 250166
Let min_count = 0, last 2 digits = 66
add minimum number of elements to min_count that sums to 66 (it is calculated from memo array
that we have precomputed).
min_count = min_count + 6,
Now, min_count = 6, last 2 digits = 01
add minimum number of elements to min_count sums to 1.
min_count = min_count + 1,
Now, min_count = 7, last 2 digits = 25
add minimum number of elements to min_count sums to 25.
min_count = min_count + 1,
Now, min_count = 8.
So, minimum number of elements chosen that sums to 250166 is 8.
Optimal chosen Elements are (250000, 100, 25, 10, 10, 10, 10, 1)

Below is the implementation of the above approach.

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `int` `minCount(``int` `K)``{``    ``// we will only store min counts``    ``// of sum upto 100``    ``int` `memo[100];` `    ``// initialize with INT_MAX``    ``for` `(``int` `i = 0; i < 100; i++) {``        ``memo[i] = INT_MAX;``    ``}` `    ``// memo[0] = 0 as 0 is``    ``// made from 0 elements``    ``memo[0] = 0;` `    ``// fill memo array with min counts``    ``// of elements that will constitute``    ``// sum upto 100` `    ``for` `(``int` `i = 1; i < 100; i++) {``        ``memo[i] = min(memo[i - 1] + 1, memo[i]);``    ``}` `    ``for` `(``int` `i = 10; i < 100; i++) {``        ``memo[i] = min(memo[i - 10] + 1, memo[i]);``    ``}` `    ``for` `(``int` `i = 25; i < 100; i++) {``        ``memo[i] = min(memo[i - 25] + 1, memo[i]);``    ``}` `    ``// min_count will store min``    ``// count of elements chosen``    ``long` `min_count = 0;` `    ``// starting from end iterate over``    ``// each 2 digits and add min count``    ``// of elements to min_count``    ``while` `(K > 0) {``        ``min_count += memo[K % 100];``        ``K /= 100;``    ``}` `    ``return` `min_count;``}` `// Driver code``int` `main()``{` `    ``int` `K = 69;` `    ``cout << minCount(K) << endl;` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach ` `class` `GFG``{``    ` `    ``static` `int` `minCount(``int` `K) ``    ``{ ``        ``// we will only store min counts ``        ``// of sum upto 100 ``        ``int` `memo[] = ``new` `int``[``100``]; ``    ` `        ``// initialize with INT_MAX ``        ``for` `(``int` `i = ``0``; i < ``100``; i++) ``        ``{ ``            ``memo[i] = Integer.MAX_VALUE; ``        ``} ``    ` `        ``// memo[0] = 0 as 0 is ``        ``// made from 0 elements ``        ``memo[``0``] = ``0``; ``    ` `        ``// fill memo array with min counts ``        ``// of elements that will constitute ``        ``// sum upto 100 ``    ` `        ``for` `(``int` `i = ``1``; i < ``100``; i++) ``        ``{ ``            ``memo[i] = Math.min(memo[i - ``1``] + ``1``, memo[i]); ``        ``} ``    ` `        ``for` `(``int` `i = ``10``; i < ``100``; i++) ``        ``{ ``            ``memo[i] = Math.min(memo[i - ``10``] + ``1``, memo[i]); ``        ``} ``    ` `        ``for` `(``int` `i = ``25``; i < ``100``; i++) ``        ``{ ``            ``memo[i] = Math.min(memo[i - ``25``] + ``1``, memo[i]); ``        ``} ``    ` `        ``// min_count will store min ``        ``// count of elements chosen ``        ``int` `min_count = ``0``; ``    ` `        ``// starting from end iterate over ``        ``// each 2 digits and add min count ``        ``// of elements to min_count ``        ``while` `(K > ``0``)``        ``{ ``            ``min_count += memo[K % ``100``]; ``            ``K /= ``100``; ``        ``} ``    ` `        ``return` `min_count; ``    ``} ``    ` `    ``// Driver code ``    ``public` `static` `void` `main (String[] args) ``    ``{``        ` `                ``int` `K = ``69``; ``    ` `                ``System.out.println(minCount(K)); ``    ``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the above approach` `def` `minCount(K):``    ` `    ``# we will only store min counts``    ``# of sum upto 100``    ``memo``=``[``10``*``*``9` `for` `i ``in` `range``(``100``)]` `    ``# memo[0] = 0 as 0 is``    ``# made from 0 elements``    ``memo[``0``] ``=` `0` `    ``# fill memo array with min counts``    ``# of elements that will constitute``    ``# sum upto 100` `    ``for` `i ``in` `range``(``1``,``100``):``        ``memo[i] ``=` `min``(memo[i ``-` `1``] ``+` `1``, memo[i])` `    ``for` `i ``in` `range``(``10``,``100``):``        ``memo[i] ``=` `min``(memo[i ``-` `10``] ``+` `1``, memo[i])` `    ``for` `i ``in` `range``(``25``,``100``):``        ``memo[i] ``=` `min``(memo[i ``-` `25``] ``+` `1``, memo[i])` `    ``# min_count will store min``    ``# count of elements chosen``    ``min_count ``=` `0` `    ``# starting from end iterate over``    ``# each 2 digits and add min count``    ``# of elements to min_count``    ``while` `(K > ``0``):``        ``min_count ``+``=` `memo[K ``%` `100``]``        ``K ``/``/``=` `100` `    ``return` `min_count` `# Driver code` `K ``=` `69` `print``(minCount(K))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of the above approach ``using` `System;` `class` `GFG``{``        ` `    ``static` `int` `minCount(``int` `K) ``    ``{ ``        ``// we will only store min counts ``        ``// of sum upto 100 ``        ``int` `[]memo = ``new` `int``[100]; ``    ` `        ``// initialize with INT_MAX ``        ``for` `(``int` `i = 0; i < 100; i++) ``        ``{ ``            ``memo[i] = ``int``.MaxValue; ``        ``} ``    ` `        ``// memo[0] = 0 as 0 is ``        ``// made from 0 elements ``        ``memo[0] = 0; ``    ` `        ``// fill memo array with min counts ``        ``// of elements that will constitute ``        ``// sum upto 100 ``    ` `        ``for` `(``int` `i = 1; i < 100; i++) ``        ``{ ``            ``memo[i] = Math.Min(memo[i - 1] + 1, memo[i]); ``        ``} ``    ` `        ``for` `(``int` `i = 10; i < 100; i++) ``        ``{ ``            ``memo[i] = Math.Min(memo[i - 10] + 1, memo[i]); ``        ``} ``    ` `        ``for` `(``int` `i = 25; i < 100; i++) ``        ``{ ``            ``memo[i] = Math.Min(memo[i - 25] + 1, memo[i]); ``        ``} ``    ` `        ``// min_count will store min ``        ``// count of elements chosen ``        ``int` `min_count = 0; ``    ` `        ``// starting from end iterate over ``        ``// each 2 digits and add min count ``        ``// of elements to min_count ``        ``while` `(K > 0)``        ``{ ``            ``min_count += memo[K % 100]; ``            ``K /= 100; ``        ``} ``    ` `        ``return` `min_count; ``    ``} ``    ` `    ``// Driver code ``    ``static` `public` `void` `Main ()``    ``{``        ` `        ``int` `K = 69; ``        ``Console.WriteLine(minCount(K)); ``    ``}``}` `// This code is contributed by ajit`

## Javascript

 ``

Output:
`9`

Time Complexity: O(1)

Auxiliary Space: O(1)

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