// 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)=>{returnMath.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);
publicstaticvoidmain(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).}}
functionarrayMin(arr){return arr.reduce(function(p, v){return( p < v ? p : v );});}functionarrayMax(arr){return arr.reduce(function(p, v){return( p > v ? p : v );});}
constfindMax=(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 ͡°)