Javascript return all indexes of highest values in an array

Here idea is

• First find the max value using math.max
• Now loop through the array and add the index if the value is same as max

let arr = [0,1,4,3,4]
let max = Math.max(...arr)

let op = []
for(let i=0; i<arr.length; i++){ 
 if(arr[i] === max){
    op.push(i)
  }
}
console.log(op)

One alternate way of doing is using reduce and a variable to keep track of max value and than access the max key's value

let arr = [0,1,4,3,4]
let max = null

let op = arr.reduce((op,inp,index)=>{
   op[inp] = op[inp] || []
   op[inp].push(index)
   if(inp > max){
    max = inp
   }
   return op
},{})
console.log(op[max])

Find the max value with reduce first, then find the indexes:

var arr = [0,1,4,3,4];

var max = arr.reduce((acc,curr) => curr > acc ? curr : acc);
var res = arr.reduce((acc,curr,idx) => curr === max ? [...acc, idx] : acc, []);

console.log(res);

Here's a slightly verbose solution that only iterates once. The idea is that you keep track of the highest value you've seen and compare against it.

let arr = [0,1,4,3,4];
let maxValue = arr[0];
let maxIndexes = [];

arr.forEach((val, index) => {
  if(maxValue === val){
    maxIndexes.push(index);
  }

  if(maxValue < val){
    maxValue = val;
    maxIndexes = [index];
  }
})

console.log(maxIndexes);

Simply, Find the max value

var max_val = Math.max.apply(null, array)

Then use reduce function

var max_val_indexes = arr.reduce(function(arr, ele, i) {
    if(ele === max_val)
        arr.push(i);
    return arr;
}, []);

To achieve expected result, use below option

1. Looping through using map
2. Capturing max index value
3. Filtering out max index values

var arr = [0,1,4,3,4]

const maxIndex = arr
  .map((v, i, self) => v === Math.max(...self) ? i : -1)
  .filter(index => index !== -1);

console.log(maxIndex)

There's no reason to find the max value and then loop again through the array. You can just keep track of the current max value as you traverse the array once:

let arr =  [0,1,4,3,4]

let maxIndexes = arr.reduce((maxes, n, i, a) => {
    let cur = a[maxes[0]]  // current max value
    if (cur == undefined || n > cur) return [i]
    return (cur == n) ? maxes.concat(i) : maxes
}, [])


console.log(maxIndexes)

Here's a fairly simple version. Once you have the maximum value, iterate over the list testing each one for a match, adding it's index if it's equal:

const allMax = (xs, max = Math.max(...xs)) => 
  xs.reduce((all, x, i) => x == max ? [...all, i] : all, [])

console.log(allMax([0, 1, 4, 3, 4]))

You could fix this up to run in a single pass (skipping the Math.max call) but the code would be more complex.

Update

This is that second version I mentioned:

const allMax = (xs) => xs.reduce(
  (a, x, i) => x > xs[a[0]] ? [i] : x < xs[a[0]] ? [...a] : [...a, i],
  [0]                                
)

console.log(allMax([0, 1, 4, 3, 4]))

This does everything in one pass. It will return [0] if you supply an empty list, which might be a problem, but it's hard to know what to do with it. One advantage is that it will work on other sorts of input. allMax(['a', 'b', 'd', 'c', 'd']) //=> [2, 4]. Dates should also work or anything you can compare with <.

And it's not as complex as I imagined.

you can get two array one if the duplicate maxim and another is the indexof the maxim number. please check the below code it might help you bro

let a = [1,3,6,6]
Math.max.apply( Math, a );
let index =[];
let y=0;
let maxValue = a.reduce( function (value,obj){
  if(Math.max.apply( Math, a) == obj){
    	index.push(y);
 		 	value.push(obj);
  }
  ++y;
  return value;
},[]);

console.log(maxValue);
console.log(index);