Problem Description
Given an integer array nums where every element appears three times except for one, which appears exactly once. Find the single element and return it.
You must implement a solution with a linear runtime complexity and use only constant extra space. leetcode
Solution
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class Solution {
public int singleNumber(int[] nums) {
int[] arr = new int[32];
for(int i=0; i<=31; i++){
int count = countSetForAtPosition(nums, i);
System.out.println(count);
if(count%3 != 0) arr[31-i] = 1;
}
StringBuffer sb = new StringBuffer();
for(int i=0; i<arr.length; i++){
sb.append(arr[i]+"");
}
long l = Long.parseLong(sb.toString(), 2);
int number = (int) l;
return number;
}
public int countSetForAtPosition(int[] nums, int position){
int c=0;
for(int i=0; i<nums.length; i++){
if(checkSetAtPosition(nums[i], position)) c++;
}
return c;
}
public boolean checkSetAtPosition(int n, int i){
n = n>>i;
return (n&1)==1;
}
}
Another way to code
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class Solution {
public int singleNumber(int[] nums) {
int ans = 0;
for(int i=0; i<=31; i++){
int count = countSetAtPosition(nums, i);
if(count%3 != 0){
ans = ans|(1<<i); //1<<i is equivalent of 2^i. We are basically setting the bit at ith position
}
}
return ans;
}
public int countSetAtPosition(int[] nums, int position){
int c=0;
for(int i=0; i<nums.length; i++){
if(checkSetAtPosition(nums[i], position)) c++;
}
return c;
}
public boolean checkSetAtPosition(int n, int i){
n = n>>i;
return (n&1)==1;
}
}