// 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);
#include <stdio.h>
int main() {
int i, n;
float arr[100];
printf("Enter the number of elements (1 to 100): ");
scanf("%d", &n);
for (i = 0; i < n; ++i) {
printf("Enter number%d: ", i + 1);
scanf("%f", &arr[i]);
}
// storing the largest number to arr[0]
for (i = 1; i < n; ++i) {
if (arr[0] < arr[i])
arr[0] = arr[i];
}
printf("Largest element = %.2f", arr[0]);
return 0;
}
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 findHighestNumber(nums) {
let inputs = nums.filter((val, i) => nums.indexOf(val) === i)
let max = Math.max(...nums);
let min = Math.min(...nums);
return max + (-min);
}
console.log(difference([1, 7, 18, -1, -2, 9]));
(function () {
const arr = [23, 65, 3, 19, 42, 74, 56, 8, 88];
function findMaxArrValue(arr) {
if (arr.length) {
let max = -Infinity;
for (let num of arr) {
max = num > max ? num : max;
}
return max;
}
return 0; // or any value what you need
}
console.log(findMaxArrValue(arr)); // => 88
})();
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 ͡°)
---without sort method---
public static int maxValue( int[] n ) {
int max = Integer.MIN_VALUE;
for(int each: n)
if(each > max)
max = each;
return max;
---with sort method---
public static int maxValue( int[] n ) {
Arrays.sort( n );
return n [ n.lenth-1 ];
}
// 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);