how to get duplicate values from array in javascript
var array = [1, 2, 2, 3, 3, 4, 5, 6, 2, 3, 7, 8, 5, 22, 1, 2, 511, 12, 50, 22];
console.log([...new Set(
array.filter((value, index, self) => self.indexOf(value) !== index))]
);
how to get duplicate values from array in javascript
var array = [1, 2, 2, 3, 3, 4, 5, 6, 2, 3, 7, 8, 5, 22, 1, 2, 511, 12, 50, 22];
console.log([...new Set(
array.filter((value, index, self) => self.indexOf(value) !== index))]
);
java array check duplicates
duplicates = false;
for(i = 0; i < zipcodeList.length; i++) {
for(j = i + 1; k < zipcodeList.length; j++) {
if(j != i && zipcodeList[j] == zipcodeList[i]) {
duplicates = true;
}
}
}
how to get duplicate values from array in javascript
var input = [1, 2, 3, 1, 3, 1];
var duplicates = input.reduce(function(acc, el, i, arr) {
if (arr.indexOf(el) !== i && acc.indexOf(el) < 0) acc.push(el); return acc;
}, []);
document.write(duplicates); // = 1,3 (actual array == [1, 3])
find duplicate in an array using xor
int DuplicateNumber(int arr[], int size){
int ans=0;
for(int i=0;i<size;i++){
ans= ans ^ arr[i] ;
}
for(int i=0;i<=size-2;i++){
ans= ans ^ i;
}
return ans;
}
Find the duplicate in an array of N integers.
// 287. Find the Duplicate Number
// Medium
// Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
// Example 1:
// Input: [1,3,4,2,2]
// Output: 2
// Example 2:
// Input: [3,1,3,4,2]
// Output: 3
// Note:
// You must not modify the array (assume the array is read only).
// You must use only constant, O(1) extra space.
// Your runtime complexity should be less than O(n2).
// There is only one duplicate number in the array, but it could be repeated more than once.
class Solution {
public:
int findDuplicate(vector<int>& nums) {
int n=nums.size();
int s=nums[0];
int f=nums[nums[0]];
while(s!=f) {
s=nums[s];
f=nums[nums[f]];
}
f=0;
while(s!=f) {
s=nums[s];
f=nums[f];
}
return s;
}
};
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