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Q:

check if number appears odd number of times in array javascript

function findOdd(A) {
    let counts = A.reduce((p, n) => (p[n] = ++p[n] || 1, p), {});
    return +Object.keys(counts).find(k => counts[k] % 2) || undefined;
}
-1
var numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12];

var evenNumbers = numbers.filter(function(item) {
   return (item % 2 == 0);
});

console.log(evenNumbers);
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///sum of odd numbers in an array javascript without loop
var numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12];

var evenNumbers = numbers.filter(function(item) {
   return (item % 2 == 0);
});

console.log(evenNumbers);
0
var total = numbers.reduce(function(total, current) {
    return total + current;
}, 0);

console.log(total);
0

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