Problem:
Given a string, return true if the number of appearances of "is" anywhere in the string is equal to the number of appearances of "not" anywhere in the string (case sensitive).
equalIsNot("This is not") → false
equalIsNot("This is notnot") → true
equalIsNot("noisxxnotyynotxisi") → true
Solution:
public boolean equalIsNot(String str) {
int len = str.length();
int not = 0;
int is = 0;
for (int i = 0; i < len; i++) {
if (i < len - 2) {
String tmp = str.substring(i,i+3);
if (tmp.equals("not"))
not++;
}
if (i < len - 1) {
String tmp2 = str.substring(i,i+2);
if (tmp2.equals("is"))
is++;
}
}
if (not == is)
return true;
else
return false;
}
Here's my solution, this seems a little less jumbled in my opinion.
ReplyDeleteint isCount = 0;
int notCount = 0;
for(int i = 0; i < str.length() - 2; i++) {
if(str.substring(i,i+3).equals("not")) notCount++;
}
for(int i = 0; i < str.length() - 1; i++) {
if(str.substring(i,i+2).equals("is")) isCount++;
}
return isCount == notCount;
Here is mine:
DeleteBefore I thought we also need to compare in uppercase/lowercase cases, but it turns out that we don't need to do it.
public boolean equalIsNot(String str) {
int count =0;
int count2=0;
for(int i =0; i<str.length();i++){
if (i+1 < str.length() && str.substring(i,i+2).equals("is")){
count++;
}
if (i+2 < str.length() && str.substring(i,i+3).equals("not")){
count2++;
}
}
return count == count2 ;
}
I do not understand why there is one testcase that did not pass. It seems the solution is logic sound.
ReplyDeleteint countIs=0;
int countNot=0;
int len=str.length();
int index=0;
while (str.indexOf("is",index+2)!=-1)
{ index=str.indexOf("is",index+2);
countIs++;
}
while (str.indexOf("not",index+3)!=-1)
{ index=str.indexOf("not",index+3);
countNot++;
}
return countIs==countNot;
}
Easy
ReplyDeletepublic boolean equalIsNot(String str) {
int isAmt = 0;
int notAmt = 0;
int incr = 0;
while(str.indexOf("is", incr) != -1)
{
isAmt++;
incr = str.indexOf("is", incr) + 1;
}
incr = 0;
while(str.indexOf("not", incr) != -1)
{
notAmt++;
incr = str.indexOf("not", incr) + 1;
}
if (isAmt == notAmt)
{
return true;
}
else
{
return false;
}
}
I shot for an easy solution that stayed away from substrings. A bit unconventional, but works.
ReplyDeletepublic static boolean equalIsNot(String str){
String strNotMissing = str.replaceAll("not", "");
String strIsMissing = str.replaceAll("is", "");
int notCount = (str.length() - strNotMissing.length()) / 3;
int isCount = (str.length() - strIsMissing.length()) / 2;
return isCount == notCount;
}
can you plz explain why /3 and /2.
Deleteint count =0;
ReplyDeletefor(int i=0;i<str.length()-1;i++)
{
if(str.substring(i,i+2).equals("is"))
{
count++;
}
}
for(int i=0;i<str.length()-2;i++)
{
if(str.substring(i,i+3).equals("not"))
{
count--;
}
}
return count==0;
public boolean equalIsNot(String str) {
ReplyDeletereturn countIt(str, "is") == countIt(str, "not");
}
public int countIt(String str, String s){
int count=0;
while(str.contains(s)){
count++;
str=str.substring(str.indexOf(s) + s.length());
}
return count;
}
public boolean equalIsNot(String str) {
ReplyDeleteint countIs = 0;
int countNot = 0;
for (int i = 0; i < str.length(); i++)
{
if (i + 2 <= str.length() && str.substring(i, i + 2).equals("is"))
{
i += 1;
countIs++;
}
else if (i + 3 <= str.length() && str.substring(i, i + 3).equals("not"))
{
i += 2;
countNot++;
}
}
return countIs == countNot;
}
public boolean equalIsNot(String str) {
ReplyDeleteint countIS = 0;
int countNOT = 0;
for(int i=0; i<str.length(); i++){
if(str.substring(i).startsWith("is")){
countIS++;
}
if(str.substring(i).startsWith("not")){
countNOT++;
}
}
return (countIS==countNOT);
}
Here is my solution:
ReplyDeletepublic boolean equalIsNot(String str) {
return str.split("is", -1).length == str.split("not", -1).length;
}
public boolean equalIsNot(String str) {
ReplyDeleteint i, b, c = 0, d = 0;
int cs1 = 0;
int cs2 = 0;
do {
i = str.indexOf("is", c);
b = str.indexOf("not", d);
if (i != -1) {
cs1++;
c = i + "is".length();
}
if (b != -1) {
cs2++;
d = b + "not".length();
}
} while (i != -1 || b != -1);
return cs1 == cs2;
}
// Simple solution using Java's .toCharArray() method
ReplyDelete// with linear time complexity of O(n);
public boolean equalIsNot(String str) {
char[] arr = str.toCharArray();
int isCount = 0;
int notCount = 0;
for (int i = 0; i < arr.length-1; i++) {
if (arr[i] == 'i' && arr[i+1] == 's') {
isCount++;
}
else if (i + 2 < arr.length &&
arr[i] == 'n' && arr[i+1] == 'o' && arr[i+2] == 't') {
notCount++;
}
}
return isCount == notCount;
}
public boolean equalIsNot(String str) {
ReplyDeleteint iscounter = 0;
int notcounter = 0;
for (int i = 0; i < str.length() - 1; i++) {
if (str.charAt(i) == 'i' && str.charAt(i + 1) == 's') {
iscounter++;
}
}
String str1 = "";
for (int i = 0; i < str.length() - 2; i++) {
str1 = str.substring(i, i + 3);
if (str1.equals("not")) {
notcounter++;
}
}
return iscounter == notcounter;
}
ReplyDeletepublic boolean equalIsNot(String str) {
char[] cArray = str.toCharArray();
int isCount = 0;
int notCount = 0;
for (int i = 0; i < cArray.length-1; i++) {
if (cArray[i] == 'i' && cArray[i + 1] == 's') {
isCount++;
}
if (cArray[i] == 'n' && cArray[i + 1] == 'o' && cArray[i + 2] == 't') {
notCount++;
}
}
if (notCount == isCount) {
return true;
} else {
return false;
}
}
public boolean equalIsNot(String str) {
ReplyDelete//str=str.toLowerCase();
str=str.replace("is","-");
str=str.replace("not","+");
int cM=0,cP=0;
for(int i=0;i<str.length();i++){
if(str.charAt(i)=='-') cM++;
if(str.charAt(i)=='+') cP++;
}
return (cM++==cP++);
}
function equalIsNot(str){
ReplyDeletevar iscount = 0
var notcount = 0
if (str.length<=1){
return true
}
for(var i=0;i<str.length;i++){
if (str.substring(i,i+1)=="i" && str.substring(i+1,i+2)=="s"){
iscount++
}
if(str.substring(i,i+1)=="n" && str.substring(i+1,i+2)=="o" && str.substring(i+2,i+3)=="t"){
notcount++
}
}
if iscount == notcount{
return true
}
else{
return false
}
}
public boolean equalIsNot(String str) {
ReplyDeleteint sum=0;
for(int i=0; i=2 && str.substring(str.length()-2).equals("is"))
sum++;
return sum==0;
}
public boolean equalIsNot(String str) {
Deleteint sum=0;
for(int i=0; i=2 && str.substring(str.length()-2).equals("is"))
sum++;
return sum==0;
}