megido
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Easy to read, simple and effective Java solution using Set:
public List<Integer> intersectArrays(int[] a, int[] b) {
Map<Integer, Integer> map = new HashMap<>(); // number of appearances
List<Integer> result = new ArrayList<>();
for (int i = 0; i < a.length; i++) {
int curNum = 1;
if (map.containsKey(a[i]))
curNum = map.get(a[i]) + 1;
map.put(a[i], curNum);
}
for (int i = 0; i < b.length; i++) {
if (!map.containsKey(b[i]))
continue;
int curNum = map.get(b[i]);
if (curNum > 0) {
curNum--;
result.add(b[i]);
}
map.put(b[i], curNum); // update decreased counter
}
return result;
}
Simple and effective Java solution using while loop:
public String charFrequency(String s) {
String res = "";
int ind = 0;
while (ind < s.length()) {
char cur = s.charAt(ind);
int num = 1;
while (ind < s.length() - 1 && cur == s.charAt(ind + 1)) {
ind++;
num++;
}
ind++;
res += String.format("%s%s", cur, num);
}
return res;
}
Effective Java solution using ArrayDeque
import java.util.*;
import java.io.*;
public class Solution {
public static void main(String[] args) {
new Solution().solve();
}
public boolean isCorrectBracketExpression(String s) {
ArrayDeque<Character> q = new ArrayDeque<>();
String brackets = "({<[)}>]";
for (Character cur : s.toCharArray()) {
if (brackets.indexOf(cur) < 4)
q.addFirst(cur);
else {
if (q.size() > 0) {
if (brackets.charAt(brackets.indexOf(q.peek()) + 4) == cur)
q.poll();
else
return false;
} else // nothing in the queue, but we want to add something, incorrect expression
return false;
}
}
return q.size() == 0;
}
public void solve() {
Scanner in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
String s = in.nextLine();
out.print(isCorrectBracketExpression(s) ? "Correct" : "Incorrect");
out.close();
}
}
Effective Java solution using HashSet
import java.util.*;
import java.io.*;
public class Solution {
public static void main(String[] args) {
new Solution().solve();
}
public void solve() {
Scanner in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int[] a = {3, 1, 4, 5, 19, 6};
int[] b = {14, 9, 22, 36, 8, 0, 64, 25};
List<Integer> result = findSquares(a, b);
for (int cur : result)
out.println(cur);
out.close();
}
// search a[i]^2 in b
public List<Integer> findSquares(int[] a, int[] b) {
List<Integer> result = new ArrayList<>();
Set<Integer> squares = new HashSet<>();
for (int i = 0; i < a.length; i++)
squares.add(a[i] * a[i]);
for (int i = 0; i < b.length; i++)
if (squares.contains(b[i]))
result.add(b[i]);
return result;
}
}
Complete Java solution with considering different edge cases
import java.util.*;
import java.io.*;
public class Solution {
/*
Smart string is a string no longer than 30 character with cut no words
*/
public static void main(String[] args) {
new Solution().solve();
}
public void solve() {
Scanner in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
String s = in.nextLine();
String res = s;
int maxLen = 10;
int firstSpace = s.indexOf(" "); // whole sentence is one word and it's longer than max allowed
if (firstSpace > maxLen)
res = "";
else if (s.length() > maxLen) {
String cutStr = s.substring(0, maxLen);
res = cutStr.substring(0, Math.max(cutStr.lastIndexOf(" "), cutStr.length()));
}
out.print(res);
out.close();
}
}
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Java solution using DFS (Depth First Search). In graph node is an index in initial array. One node is connected to some other one if the last letter equals to first letter of the second word. First we build a graph, and then run DFS starting from each node. If at some point we visit all nodes while traversing the graph, it means we found the solution.
- megido December 08, 2015