Simple Recursion in Javascript with/without a subroutine -

i'm stuck on how find max value of array recursively in javascript. tried first iteratively , of course works. recursively, wanted try subroutine first, , without subroutine. what's best way call subroutine inside itself? getting tripped want inside indices, subroutine accepts array parameter.

function maxvalue(arr) {   var max = 0;   (var = 0; < arr.length; i++) {     if (arr[i] > max) {       max = arr[i];     }   }   return max; }  console.log(maxvalue([2, 7, 8, 3, 1, 4])); //returns 8  function maxvaluerecursive(arr) {   var length = arr.length; //6   var max = 0;     function dosearch(arr) {     if (arr.length === 1) {       max = arr[0];     } else { // true        max = math.max(arr[length - 1], arr[length - 2]);       //max = math.max(4, dosearch()) = 4.      }     return max;   }    return dosearch(arr); }  console.log(maxvaluerecursive([2, 7, 8, 3, 1, 4])) //returns 4 

you can use math.max , solve smaller bit of array @ each step. trick remove (any) item array , use math.max compare item findmax on smaller array:

function findmax(arr){     if (arr.length == 1){         // base case - single item in array max         return arr[0];     }     // recursive call on smaller problem (shorter array)     return math.max(arr[0], findmax(arr.slice(1))) } 

i used slice can use pop or whatever method remove item , compare using math.max did.

for array [1, 4, 2, 3] recursion unfolds follows:

1. findmax([1, 4, 2, 3] 2. math.max(1, findmax([4, 2, 3])) 3. math.max(1, math.max(4, findmax([2, 3]))) 4. math.max(1, math.max(4, math.max(2, findmax([3])))) 5. math.max(1, math.max(4, math.max(2, 3))) // 4