# Java iterative solution with explanation, beat 75%

• ``````Base case: n = 0 print "1"
for n = 1, look at previous string and write number of times a digit is seen and the digit itself. In this case,
digit 1 is seen 1 time in a row... so print "1 1"
for n = 2, digit 1 is seen two times in a row, so print "2 1"
for n = 3, digit 2 is seen 1 time and then digit 1 is seen 1 so print "1 2 1 1"
for n = 4 you will print "1 1 1 2 2 1"
``````
``````    public String countAndSay(int n) {
StringBuilder sb = new StringBuilder();
sb.append('1');
for(int i = 1; i < n; i++){
char[] numbers = sb.toString().toCharArray();
char pre = numbers[0];

sb = new StringBuilder();
int j = 0;
while(j < numbers.length ){
int cnt = 0;
while(j<numbers.length && pre == numbers[j] ){
cnt++;
j++;
}
if(cnt>0){
sb.append(cnt);
sb.append(numbers[j-1]);
if(j < numbers.length){
pre = numbers[j];
}
}
}
}
return sb.toString();
}
``````

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