Sum Of Odd Numbers In An Array Javascript

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Sum of Array of Odd numbers - JS

1
3     5
7     9    11
13    15    17    19
21    23    25    27    29
rowSumOddNumbers(1); // 1
rowSumOddNumbers(2); // 3 + 5 = 8
function rowSumOddNumbers(n){
  let result = [];

  // generate the arrays of odd numbers
  for(let i = 0; i < 30; i++){
    // generate sub arrays by using another for loop
    // and only pushing if the length is equal to current j
    let sub = [];
    for(let j = 1; j <= n; j++){
      // if length === j (from 1 - n) keep pushing
       if(sub[j - 1].length <= j){
         // and if i is odd
         if(i % 2 !== 0){
           // push the i to sub (per length)
           sub.push(i);
         }
       }
    }
    // push everything to the main array
    result.push(sub);
  }

  // return sum of n 
  return result[n + 1].reduce(function(total, item){
    return total += item;
  });
}
1: 1 === 1 ^ 3
2: 3 + 5 = 8 === 2 ^ 3
3: 7 + 9 + 11 = 27 === 3 ^ 3
... etc
function rowSumOddNumbers(n) {
    // how many numbers are there in the rows above n?
    // sum of arithmetic sequence...
    let numbers_before_n_count = (n - 1) * n / 2;

    let first_number_in_nth_row = numbers_before_n_count * 2 + 1;
    let last_number_in_nth_row = first_number_in_nth_row + 2 * (n - 1);

    // sum of arithmetic sequence again...
    return n * (first_number_in_nth_row + last_number_in_nth_row) / 2;
}

Sum of array odd number javascript

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

console.log(total);
                                    var array=[2,3,5,6,8,7,9]
var sum =0
for(var i=0; i&lt;array.length; i++){
if(array[i]===0){
sum+=array[i]
console.log(sum)
}
}
                                    ///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);

Sum of odd numbers in an array javascript without loop

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

console.log(total);
                                    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);
                                    ///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);

Sum of odd numbers in an array javascript without loop

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

console.log(total);

Program to print Sum of even and odd elements in an array

Input : arr = {1, 2, 3, 4, 5, 6}
Output :Even index positions sum 9
        Odd index positions sum 12
Explanation: Here, n = 6 so there will be 3 even 
index positions and 3 odd index positions in an array
Even = 1 + 3 + 5 = 9
Odd =  2 + 4 + 6 = 12

Input : arr = {10, 20, 30, 40, 50, 60, 70}
Output : Even index positions sum 160
        Odd index positions sum 170
Explanation: Here, n = 7 so there will be 3 odd
index positions and 4 even index positions in an array
Even = 10 + 30 + 50 + 70 = 160
Odd = 20 + 40 + 60 = 120 
Even index positions sum 9
Odd index positions sum 12
Even index positions sum 9
Odd index positions sum 12

Check if its possible to make sum of the array odd with given Operations

Yes

JavaScript program to find the sum of all odd numbers below one given number

function isOdd(n) {
  return Boolean(n % 2);
}

function findSum(no) {
  let sum = 0;

  for (var i = 0; i < no; i++) {
    if (isOdd(i)) {
      sum += i;
    }
  }
  return sum;
}

console.log(findSum(100));

Find the Odd Numbers in an Array using JavaScript

Copied!const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];

const odds = arr.filter(number => {
  return number % 2 !== 0;
});

console.log(odds); // 👉️ [1, 3, 5, 7, 9]
Copied!const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];

const odds = [];

arr.forEach(number => {
  if (number % 2 !== 0) {
    odds.push(number);
  }
});

console.log(odds); // 👉️ [1, 3, 5, 7, 9]

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