// For large data, it's better to use reduce. Supose arr has a large data in this case:
const arr = [1, 5, 3, 5, 2];
const max = arr.reduce((a, b) => { return Math.max(a, b) });
// For arrays with relatively few elements you can use apply:
const max = Math.max.apply(null, arr);
// or spread operator:
const max = Math.max(...arr);
public static void main(String[] args) {
int[] xr = {2, 4, 1, 3, 7, 5, 6, 10, 8, 9};
//find maximum value
int max = xr[0];
for (int i = 0; i < xr.length; i++) {
if (xr[i] > max) {
max = xr[i];
}
}
//find minimum value
int min=xr[0];
for (int i = 0; i <xr.length ; i++) {
if (xr[i]<min){
min=xr[i];
}
}
System.out.println("max: "+max);
System.out.println("min: "+min);
}
// For regular arrays:
var max = Math.max(...arrayOfNumbers);
// For arrays with tens of thousands of items:
let max = testArray[0]; //here we have considered max to the first element because we don't know which is max yet.
for (let i = 1; i < testArrayLength; ++i) {
if (testArray[i] > max) { //in each iteration it will compare if the value is greater than the current considered value (we just considered first element)
max = testArray[i]; //in the above iteration if the testArray find value/element greater than max then this new max value will be considered as Max (this will happen until the max value found).
}
}
function arrayMin(arr) {
return arr.reduce(function (p, v) {
return ( p < v ? p : v );
});
}
function arrayMax(arr) {
return arr.reduce(function (p, v) {
return ( p > v ? p : v );
});
}
const findMax = (arr)=>{
let max = 0 ;
for (let index = 0; index < arr.length; index++) {
if (max < arr[index] && max != arr[index]) {
max = arr[index];
}
}
return max;
}
//if you find this answer is useful ,
//upvote ⇑⇑ , so can the others benefit also . @mohammad alshraideh ( ͡~ ͜ʖ ͡°)