71 lines
2.1 KiB
C++
71 lines
2.1 KiB
C++
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// Source : https://leetcode.com/problems/maximum-number-of-balls-in-a-box/
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// Author : Hao Chen
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// Date : 2021-03-27
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/*****************************************************************************************************
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*
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* You are working in a ball factory where you have n balls numbered from lowLimit up to highLimit
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* inclusive (i.e., n == highLimit - lowLimit + 1), and an infinite number of boxes numbered from 1 to
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* infinity.
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*
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* Your job at this factory is to put each ball in the box with a number equal to the sum of digits of
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* the ball's number. For example, the ball number 321 will be put in the box number 3 + 2 + 1 = 6 and
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* the ball number 10 will be put in the box number 1 + 0 = 1.
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*
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* Given two integers lowLimit and highLimit, return the number of balls in the box with the most
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* balls.
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*
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* Example 1:
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*
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* Input: lowLimit = 1, highLimit = 10
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* Output: 2
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* Explanation:
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* Box Number: 1 2 3 4 5 6 7 8 9 10 11 ...
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* Ball Count: 2 1 1 1 1 1 1 1 1 0 0 ...
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* Box 1 has the most number of balls with 2 balls.
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*
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* Example 2:
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*
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* Input: lowLimit = 5, highLimit = 15
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* Output: 2
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* Explanation:
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* Box Number: 1 2 3 4 5 6 7 8 9 10 11 ...
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* Ball Count: 1 1 1 1 2 2 1 1 1 0 0 ...
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* Boxes 5 and 6 have the most number of balls with 2 balls in each.
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*
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* Example 3:
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*
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* Input: lowLimit = 19, highLimit = 28
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* Output: 2
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* Explanation:
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* Box Number: 1 2 3 4 5 6 7 8 9 10 11 12 ...
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* Ball Count: 0 1 1 1 1 1 1 1 1 2 0 0 ...
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* Box 10 has the most number of balls with 2 balls.
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*
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* Constraints:
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*
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* 1 <= lowLimit <= highLimit <= 10^5
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******************************************************************************************************/
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class Solution {
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private:
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int sum(int n) {
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int s = 0;
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for(; n > 0; n /= 10){
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s += n % 10;
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}
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return s;
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}
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public:
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int countBalls(int lowLimit, int highLimit) {
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int cnt[46] ={0}; //10^5 means 9+9+9+9+9 = 45
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int m = 0;
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for (int n = lowLimit; n<=highLimit; n++) {
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int box = sum(n);
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cnt[box]++;
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m = max(m, cnt[box]);
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}
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return m;
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}
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};
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