[Microsoft Google interview question 100-[45] an integer array with a length of N, which is divided into M parts, so that the sum of each part is equal, and the maximum value of M is obtained.

45th questions

An integer array with the length of N. It is divided into M parts so that the sum of the parts is equal and the maximum value of M is obtained.

For example:

{3, 2, 4, 3, 6} can be divided:

{3, 2, 4, 3, 6} M = 1;

{3, 6} {2, 4, 3} m = 2

{3, 3} {2, 4} {6} M = 3 so the maximum m value is 3

Since the sum of parts is equal, the first thing that comes to mind is to sum up the array and then solve the number of groups.

Then define the variableI = length (array length), I> 0, I --To find the most satisfyingI

import java.util.ArrayList;import java.util.List;public class Algorithm45_0 {static int getMax(int a[]){int sum=0;List<Integer> myList=new ArrayList<Integer>();for(int aa:a){sum+=aa;myList.add(aa);}for(int i=a.length;i>0;i--){if(sum%i==0){if(ifExsits(myList,sum/i,sum/i)){return i;}}}return 1;}static boolean ifExsits(List<Integer> rootList,int num,int orgNum){if(rootList.size()==0&&num==orgNum){return true;}boolean flag=false;for(int i=0;i<rootList.size();i++){List<Integer> subList=new ArrayList<Integer>();for(int j=0;j<rootList.size();j++){if(j!=i){subList.add(rootList.get(j));}}if(rootList.get(i)==num){flag=flag||ifExsits(subList, orgNum, orgNum);}else if(rootList.get(i)<num){flag=flag||ifExsits(subList, num-rootList.get(i), orgNum);}}return flag;}public static void main(String[] args) {// TODO Auto-generated method stubint a[]={1,2,3,6,6};int b=getMax(a);System.err.println(b);}}