# Your question: How do you find the lowest value in an array?

Contents

For an array of ascending order the first element is the smallest element, you can get it by arr[0] (0 based indexing). If the array is sorted in descending order then the last element is the smallest element,you can get it by arr[sizeOfArray-1].

## How do you find the minimum value of an array?

Algorithm

1. STEP 1: START.
2. STEP 2: INITIALIZE arr[] = {25, 11, 7, 75, 56}
3. STEP 3: min = arr[0]
4. STEP 4: REPEAT STEP 5 for(i=0; i
5. STEP 5: if(arr[i]min) min=arr[i]
6. STEP 6: PRINT “Smallest element in given array:”
7. STEP 7: PRINT min.
8. STEP 8: END.

## How do you find the maximum and minimum of an array?

Logic to find maximum and minimum element in array

Assume first array element as maximum and minimum both, say max = arr[0] and min = arr[0] . Iterate through array to find maximum and minimum element in array. Run loop from first to last array element i.e. 0 to size – 1 .

## How do you find the minimum value?

If you have the equation in the form of y = ax^2 + bx + c, then you can find the minimum value using the equation min = c – b^2/4a. If you have the equation y = a(x – h)^2 + k and the a term is positive, then the minimum value will be the value of k.

## How do you find the minimum and second of an array?

Let’s see the full example to find the second smallest number in java array.

1. public class SecondSmallestInArrayExample{
2. public static int getSecondSmallest(int[] a, int total){
3. int temp;
4. for (int i = 0; i
5. {
6. for (int j = i + 1; j
7. {
8. if (a[i] > a[j])

## How do you find the maximum value in an array?

Java program to find largest number in an array

1. Initialize array value.
2. Step 2: (int max = a[0];) Initialize max value as array’s first value.
3. Step 3: (for int i = 1; i
4. Step 4: if(a[i] > max)
5. Continue the loop and resign the max value if array current value is greater than max.
6. Program. …
7. Output.

11 мар. 2016 г.

## How do you find the maximum and minimum of an array in Python?

Python provides different inbuilt function. min() is used for find out minimum value in an array, max() is used for find out maximum value in an array. index() is used for finding the index of the element.

## How do you find the maximum and minimum of an array in C++?

The program output is shown below.

1. #include
2. int arr[10], n, i, max, min;
3. cout array : “;
4. cin >> n;
5. cout array : “;
6. for (i = 0; i
7. cin >> arr[i];
8. max = arr[0];

## How many comparisons are required to find the maximum and minimum?

Ans: minimum number of comparison require to find minimum and maximum is: Approach is divide and conquer …. Thus, the approach does (3/2)n−2 comparisons if n is a power of 2. And it does more than (3/2)n−2 comparisons if n is not a power of 2.

## How many comparisons are needed to find the maximum value in an array N?

Solution Steps

Return max and min. If n is a power of 2, the algorithm needs exactly 3n/2–2 comparisons to find min and max.

## What is the minimum in math?

Minimum, in mathematics, point at which the value of a function is less than or equal to the value at any nearby point (local minimum) or at any point (absolute minimum); see extremum. …

## How do you find the first two minimum values of an array?

A Simple Solution is to sort the array in increasing order. The first two elements in sorted array would be two smallest elements. Time complexity of this solution is O(n Log n). A Better Solution is to scan the array twice.

## How do you find the second largest number in an array?

Find 2nd Largest Number in Array using Arrays

1. import java.util.Arrays;
2. public class SecondLargestInArrayExample1{
3. public static int getSecondLargest(int[] a, int total){
4. Arrays.sort(a);
5. return a[total-2];
6. }
7. public static void main(String args[]){
8. int a[]={1,2,5,6,3,2};

## How do you find the lowest number in an array in C++?

Algorithm to find smallest element of array

Initialize one variables minElement with first element of inputArray. Using a loop, traverse inputArray from index 0 to N -1 and compare each element with minElement. If current element is less than minElement, then update minElement with current element.