java - Storing hashmap values as program iterates -

i trying solve programming challenge outlined below, wherein need find word in given sentence greatest number of repeated characters. have struggled bit this, , lucky enough find code counting occurrences of letters in string (also below). particular piece of code store letters in hashmap, , need tailor stores character occurrence of each word separately (instead of in aggregate, doing @ moment). stuck. use store state of hashmap each iteration of loop?

/* using java language, have function lettercounti(str) take   * str parameter being passed , return first word   * greatest number of repeated letters. example: "today,   * greatest day ever!" should return greatest because has 2 e's   * (and 2 t's) , comes before ever has 2 e's. if there   * no words repeating letters return -1. words   * separated spaces. */  import java.util.arraylist; import java.util.hashmap; import java.util.map;  public class othercountletters {     void countletters2(string str) {         string[] words = str.tolowercase().split(" ");         map<character, integer> numchars = new hashmap<character, integer>();          (int = 0; < words.length; i++) {             (int j = 0; j < words[i].length(); j++) {                 char charat = words[i].charat(j);                  if (!numchars.containskey(charat)) {                     numchars.put(charat, 1);                 } else {                     numchars.put(charat, numchars.get(charat) + 1);                 }             }          }          system.out.println(numchars);     }      public static void main(string[] args) {         othercountletters ocl = new othercountletters();         ocl.countletters2("today greatest day ever");     }  } 

at moment, sentence "today greatest day ever", program returns

{v=1, g=1, d=2, e=5, t=4, s=2, r=2, a=3, o=1, h=1, y=2, i=1} 

but need return like

{a=1, d=1, o=1, t=1, y=1} //'today' {i=1, s=1}                //'is' {e=1, h=1, t=1}           //'the' {g=1, t=2, e=2, s=1, r=1, a=1} //'greatest' {d=1, a=1, y=1}           //'day' {v=1, e=2, r=1}           //'ever' 

that way, iterate on each entry see 1 has largest value, , return corresponding word user.

thanks,

-----edit----

after posting had eureka moment:

import java.util.arraylist; import java.util.hashmap; import java.util.map;  public class othercountletters {     void countletters2(string str) {         string[] words = str.tolowercase().split(" ");         string target = null;         int largest = 0;         map<character, integer> numchars = new hashmap<character, integer>();          (int = 0; < words.length; i++) {             (int j = 0; j < words[i].length(); j++) {                 char charat = words[i].charat(j);                  if (!numchars.containskey(charat)) {                     numchars.put(charat, 1);                 } else {                     numchars.put(charat, numchars.get(charat) + 1);                 }                 if (numchars.get(charat) > largest) {                     largest = numchars.get(charat);                     target = words[i];                 }             }             numchars.clear();         }         if (largest != 1) {             system.out.println(target);         } else {             system.out.println("there no words 2 or more letters");         }     }      public static void main(string[] args) {         othercountletters ocl = new othercountletters();         ocl.countletters2("today greatest day ever , car");     }  } 

have consider split phrase word , iterate on every word using method 'othercountletters'?

instead of returning nothing, return highest score of repeated character. then, in loop , have compare current maximum

at end of loop, might able give word of phrase highest repeated character