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By : Aleksander Drozd
Date : November 20 2020, 03:01 PM
it should still fix some issue The issue is integer division.
In Java when you divide two integers, you get an integer (losing anything past the decimal point). code : ## How to find the biggest number in an array

By : Иван Литвинчук
Date : March 29 2020, 07:55 AM
hop of those help? Iterate over the array until you hit your break condition.
While iterating (use the for or while loop), remember the highest value so far and compare against current. ## How to find the biggest number in array

By : Michael Janea
Date : March 29 2020, 07:55 AM
With these it helps How can I find the biggest number from the numbers I've entered in array? , You're very close:
code :
``````Math.max.apply(Math, number);
`````` ## How to find the biggest, second biggest and third biggest number in an array, then display their sequence location?

By : Yana Mikhnich
Date : March 29 2020, 07:55 AM
it helps some times this block might be problematic because you're repurposing high, sec, and thr from representing the values of the array to representing the index of the array.
Not only that, but you're depending on high, sec, and thr, being values of the array throughout the loop.
code :
``````for (a = 0; a < dval.length; a++) // Determines sequence location of first second and third highest numbers
{
if (dval[a] == high)
{
high = a+1;
}
if (dval[a] == sec)
{
sec = a+1;
}
if (dval[a] == thr)
{
thr = a+1;
}
}
`````` ## What is the fastest way to find Nth biggest number of an INT array?

By : hnm
Date : March 29 2020, 07:55 AM
should help you out Randomized quickselect algorithm works in average case complexity O(n). Practically it's very rare to be O(n^2). It uses quicksort's partition function ## Pythonic Way to Find Next Biggest Number in an Array?

By : Alvin
Date : March 29 2020, 07:55 AM
I wish this helpful for you One possible solution (for small datasets) is to find the minimum value of all the values in Max's that are greater than each value in Closes:
code :
``````closes = [1, 2, 3, 4, 5, 4, 3, 2, 1]
maxs = [3, 5, 6, 7, 2]
nextbig = [min([m for m in maxs if m > c]) for c in closes]
print(nextbig)
``````
``````[2, 3, 5, 5, 6, 5, 5, 3, 2]
``````
``````closes = [1, 2, 3, 4, 5, 4, 3, 2, 1]
maxs = [3, 5, 6, 7, 2]
nextbig = [c for c in closes]
maxs.sort()
for m in maxs:
for i, c in enumerate(closes):
if m > c and nextbig[i] == c:
nextbig[i] = m
print(nextbig)
``````
``````[2, 3, 5, 5, 6, 5, 5, 3, 2]
`````` 