A Test array has the following property:
a[0] = a[1] + a[2] = a[3] + a[4] + a[5] = a[6] + a[7] + a[8] + a[9] = ...
The length of a Test array must be n*(n+1)/2 for some n.
Write a method named isTestArray that returns 1 if its array argument is a Test array, otherwise it returns 0. The function signature is:
int isMadhavArray(int[ ] a)
Example:
This is what I have tried:
public class TestArray {
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println(isTestArray(new int[] {2,1,1,4,-1,-1}));
}
static int isTestArray(int[] a){
boolean isEq=true;
for(int i=0;i<a.length;i++){
int n=i, value=a.length;
int equation=n*(n+1)/2;
if(value==equation)
{
for(int x=0,y=1;y<a.length;x++,y++){
if(a[0]==a[x]+a[y]){
//having problem over here :(
}
}
}
else
isEq=false;
}
if(isEq)
return 1;
else
return 0;
}
}
Number of elements in sum increases by 1 in every portion.
You should do 2 things:
Write a method that will calculate sum of N elements starting from K position in array
Increment i with the number of the smallest portion in iteration
example code:
int portionSize = 1; // number of elements to sum
int position = 0; // index of first element
while (position + 2 * portionSize + 1 < array.length) { // condition for last iteration
if (sum(position, portionSize) != sum(position + portionSize, portionSize + 1) {
return false; // if not equal, return immediately
}
position += portionSize;
portionSize++;
}
Here is my attempt for your troubling inner for loop. Because each time you loop, you increment the number of indexes you need to add by one, so I think you need another nested loop like this. I know it is ugly. Can't blame me, I am a beginner.
public class TestArray {
public static void main(String[] args) {
System.out.println(isTestArray(new int[] {2,1,1,4,-1,-1}));
}
static int isTestArray(int[] a) {
boolean isEq = false;
for(int i=0; i < a.length; i++) {
int n=i, value = a.length;
int equation= n*(n+1)/2;
if (value==equation) {
int index = 1;
for (int x = 1; x < a.length; x++) {
int total = 0;
for (int y = index, count = 0; count < x+1; y++, count++) {
if (y!= a.length) {
total += a[y];
if (total == a[0])
isEq = true;
else
isEq = false;
index++;
}
else
break;
}
}
}
}
if (isEq)
return 1;
else
return 0;
}
public class TestArray {
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println(isTestArray(new int[] {2,1,1}));
System.out.println(isTestArray(new int[] {3, 1, 2, 3, 0}));
}
static int isTestArray(int[] a){
boolean isEq=false, isEx=false;
int equation=0,value=a.length;
//for equation check
for(int n=0;n<value;n++){
equation=n*(n+1)/2;
if(value==equation){
isEx=true;
break;
}
}
//if equation is true
if(isEx)
{
int x=1,y=3,c=0,sum=0;
do{
for(int I=x,J=y;I<J;I++){
sum=sum+a[I];
}
if(a[0]==sum){
isEq=true;
sum=0;
}
else{
isEq=false;
break;
}
c++;
x=y;
y=x+2+c;
}while(x<a.length);
}
//last operation of return
if(isEq)
return 1;
else
return 0;
}
}
import java.io.*;
public class StackQues1{
static boolean flag = true;
static int arrLength = 0;
public static void main(String args[]){
Console c = System.console();
System.out.println("Enter the length of array - ");
arrLength = Integer.parseInt(c.readLine());
//Let max val of n<=20
//To check n(n+1)/2
boolean nFlag = false;
for(int n=1;n<=20;n++){
if(arrLength == (n*(n+1))/2){
nFlag = true;
}
}
if(!nFlag){
System.out.println("Length of array is not in the form of (n*(n+1))/2");
}else{
int arr[] = new int[arrLength];
System.out.println("Enter the elements of array - ");
for(int i=0;i<arr.length;i++){
arr[i] = Integer.parseInt(c.readLine());
}
System.out.println("Returned value = "+isTestArray(arr));
}
}
//Logic Implementation
public static int isTestArray(int[] arr){
int sum = 0;
int noOfElements=0;
for(int i=0;i<arr.length;){
noOfElements++;
sum=0;
for(int j=1;j<=noOfElements;j++){
sum=sum+arr[i];
i++;
}
if(arr[0] == sum){
flag = true;
}else{
flag = false;
break;
}
}
if(flag){
return 1;
}else{
return 0;
}
}
}
I have an array of integers, and I need to find the one that's closest to zero (positive integers take priority over negative ones.)
Here is the code I have so far:
public class CloseToZero {
public static void main(String[] args) {
int[] data = {2,3,-2};
int curr = 0;
int near = data[0];
// find the element nearest to zero
for ( int i=0; i < data.length; i++ ){
curr = data[i] * data[i];
if ( curr <= (near * near) ) {
near = data[i];
}
}
System.out.println( near );
}
}
Currently I'm getting a result of -2 but I should be getting 2. What am I doing wrong?
This will do it in O(n) time:
int[] arr = {1,4,5,6,7,-1};
int closestIndex = 0;
int diff = Integer.MAX_VALUE;
for (int i = 0; i < arr.length; ++i) {
int abs = Math.abs(arr[i]);
if (abs < diff) {
closestIndex = i;
diff = abs;
} else if (abs == diff && arr[i] > 0 && arr[closestIndex] < 0) {
//same distance to zero but positive
closestIndex =i;
}
}
System.out.println(arr[closestIndex ]);
If you are using java8:
import static java.lang.Math.abs;
import static java.lang.Math.max;
public class CloseToZero {
public static void main(String[] args) {
int[] str = {2,3,-2};
Arrays.stream(str).filter(i -> i != 0)
.reduce((a, b) -> abs(a) < abs(b) ? a : (abs(a) == abs(b) ? max(a, b) : b))
.ifPresent(System.out::println);
}
}
Sort the array (add one line of code) so the last number you pick up will be positive if the same absolute value is selected for a positive and negative numbers with the same distance.
Source code:
import java.util.Arrays;
public class CloseToZero {
public static void main(String[] args) {
int[] data = {2,3,-2};
int curr = 0;
int near = data[0];
Arrays.sort(data); // add this
System.out.println(Arrays.toString(data));
// find the element nearest to zero
for ( int i=0; i < data.length; i++ ){
System.out.println("dist from " + data[i] + " = " + Math.abs(0 -data[i]));
curr = data[i] * data[i];
if ( curr <= (near * near) ) {
near = data[i];
}
}
System.out.println( near );
}
}
Just add zero to this list.
Then sort the list
Arrays.sort(data);
then grab the number before or after the zero and pick the minimum one greater than zero
Assumption is that the array data has at least 1 value.
int closestToZero = 0;
for ( int i = 1; i < data.length; i++ )
{
if ( Math.abs(data[i]) < Math.abs(data[closestToZero]) ) closestToZero = i;
}
The value in closestToZero is the index of the value closest to zero, not the value itself.
static int Solve(int N, int[] A){
int min = A[0];
for (int i=1; i<N ; i++){
min = min > Math.abs(0- A[i]) ? Math.abs(0- A[i]) : Math.abs(min);
}
return min;
}
As you multiply data[i] with data[i], a value negative and a value positive will have the same impact.
For example, in your example: 2 and -2 will be 4. So, your code is not able to sort as you need.
So, here, it takes -2 as the near value since it has the same "weight" as 2.
I have same answer with different method,Using Collections and abs , we can solved.
static int Solve(int N, int[] A){
List<Integer> mInt=new ArrayList<>();
for ( int i=0; i < A.length; i++ ){
mInt.add(Math.abs(0 -A[i]));
}
return Collections.min(mInt);
}
That all,As simple as that
This is a very easy to read O(n) solution for this problem.
int bigestNegative = Integer.MIN_VALUE;
int smalestpositive = Integer.MAX_VALUE;
int result = 0;
for (int i = 0; i < n; i++) {
//if the zero should be considered as result as well
if ( temperatures[i] == 0 ) {
result = 0;
break;
}
if ( temperatures[i] > 0 && temperatures[i] < smalestpositive ) {
smalestpositive = temperatures[i];
}
if ( temperatures[i] < 0 && temperatures[i] > bigestNegative ) {
bigestNegative = temperatures[i];
}
}
if( (Math.abs(bigestNegative)) < (Math.abs(smalestpositive)) && bigestNegative != Integer.MIN_VALUE)
result = bigestNegative;
else
result = smalestpositive;
System.out.println( result );
First convert the int array into stream. Then sort it with default sorting order. Then filter greater than zero & peek the first element & print it.
Do it in declarative style which describes 'what to do', not 'how to do'. This style is more readable.
int[] data = {2,3,-2};
IntStream.of(data)
.filter(i -> i>0)
.sorted()
.limit(1)
.forEach(System.out::println);
using Set Collection and abs methode to avoid complex algo
public static void main(String[] args) {
int [] temperature={0};
***// will erase double values and order them from small to big***
Set<Integer> s= new HashSet<Integer>();
if (temperature.length!=0) {
for(int i=0; i<temperature.length; i++) {
***// push the abs value to the set***
s.add(Math.abs(temperature[i]));
}
// remove a zero if exists in the set
while(s.contains(0)) {
s.remove(0);
}
***// get first (smallest) element of the set : by default it is sorted***
if (s.size()!=0) {
Iterator iter = s.iterator();
System.out.println(iter.next());
}
else System.out.println(0);
}
else System.out.println(0);
}
static int nearToZero(int[] A){
Arrays.sort(A);
int ans = 0;
List<Integer> list = Arrays.stream(A).boxed().collect(Collectors.toList());
List<Integer> toRemove = new ArrayList<>();
List<Integer> newList = new ArrayList<>();
for(int num: list){
if(newList.contains(num)) toRemove.add(num);
else newList.add(num);
}
list.removeAll(toRemove);
for(int num : list){
if(num == 0 ) return 0;
if(ans == 0 )ans = num;
if(num < 0 && ans < num) ans = num;
if(num < ans) ans = num;
if(num > 0 && Math.abs(ans) >= num) ans = num;
}
return ans;
}
here is a method that gives you the nearest to zero.
use case 1 : {1,3,-2} ==> return 1 : use the Math.abs() for comparison and get the least.
use case 2 : {2,3,-2} ==> return 2 : use the Math.abs() for comparison and get the Math.abs(least)
use case 3 : {-2,3,-2} ==> return -2: use the Math.abs() for comparison and get the least.
public static double getClosestToZero(double[] liste) {
// if the list is empty return 0
if (liste.length != 0) {
double near = liste[0];
for (int i = 0; i < liste.length; i++) {
// here we are using Math.abs to manage the negative and
// positive number
if (Math.abs(liste[i]) <= Math.abs(near)) {
// manage the case when we have two equal neagative numbers
if (liste[i] == -near) {
near = Math.abs(liste[i]);
} else {
near = liste[i];
}
}
}
return near;
} else {
return 0;
}
}
You can do like this:
String res = "";
Arrays.sort(arr);
int num = arr[0];
int ClosestValue = 0;
for (int i = 0; i < arr.length; i++)
{
//for negatives
if (arr[i] < ClosestValue && arr[i] > num)
num = arr[i];
//for positives
if (arr[i] > ClosestValue && num < ClosestValue)
num = arr[i];
}
res = num;
System.out.println(res);
First of all you need to store all your numbers into an array. After that sort the array --> that's the trick who will make you don't use Math.abs(). Now is time to make a loop that iterates through the array. Knowing that array is sorted is important that you start to make first an IF statement for negatives numbers then for the positives (in this way if you will have two values closest to zero, let suppose -1 and 1 --> will print the positive one).
Hope this will help you.
The easiest way to deal with this is split the array into positive and negative sort and push the first two items from both the arrays into another array. Have fun!
function closeToZeroTwo(arr){
let arrNeg = arr.filter(x => x < 0).sort();
let arrPos = arr.filter(x => x > 0).sort();
let retArr = [];
retArr.push(arrNeg[0], arrPos[0]);
console.log(retArr)
}
Easiest way to just sort that array in ascending order suppose input is like :
int[] array = {10,-5,5,2,7,-4,28,65,95,85,12,45};
then after sorting it will gives output like:
{-5,-4,2,5,7,10,12,28,45,65,85,95,}
and for positive integer number, the Closest Positive number is: 2
Logic :
public class Closest {
public static int getClosestToZero(int[] a) {
int temp=0;
//following for is used for sorting an array in ascending nubmer
for (int i = 0; i < a.length-1; i++) {
for (int j = 0; j < a.length-i-1; j++) {
if (a[j]>a[j+1]) {
temp = a[j];
a[j]=a[j+1];
a[j+1]=temp;
}
}
}
//to check sorted array with negative & positive values
System.out.print("{");
for(int number:a)
System.out.print(number + ",");
System.out.print("}\n");
//logic for check closest positive and Integer
for (int i = 0; i < a.length; i++) {
if (a[i]<0 && a[i+1]>0) {
temp = a[i+1];
}
}
return temp;
}
public static void main(String[] args) {
int[] array = {10,-5,5,2,7,-4,28,65,95,85,12,45};
int closets =getClosestToZero(array);
System.out.println("The Closest Positive number is : "+closets);
}
}
static void closestToZero(){
int[] arr = {45,-4,-12,-2,7,4};
int max = Integer.MAX_VALUE;
int closest = 0;
for (int i = 0; i < arr.length; i++){
int value = arr[i];
int abs = Math.abs(value);
if (abs < max){
max = abs;
closest = value;
}else if (abs == max){
if (value > closest){
closest = value;
}
}
}
Return a positive integer if two absolute values are the same.
package solution;
import java.util.Scanner;
public class Solution {
public static void trier(int tab[]) {
int tmp = 0;
for(int i = 0; i < (tab.length - 1); i++) {
for(int j = (i+1); j< tab.length; j++) {
if(tab[i] > tab[j]) {
tmp = tab[i];
tab[i] = tab[j];
tab[j] = tmp;
}
}
}
int prochePositif = TableauPositif(tab);
int procheNegatif = TableauNegatif(tab);
System.out.println(distanceDeZero(procheNegatif,prochePositif));
}
public static int TableauNegatif(int tab[]) {
int taille = TailleNegatif(tab);
int tabNegatif[] = new int[taille];
for(int i = 0; i< tabNegatif.length; i++) {
tabNegatif[i] = tab[i];
}
int max = tabNegatif[0];
for(int i = 0; i <tabNegatif.length; i++) {
if(max < tabNegatif[i])
max = tabNegatif[i];
}
return max;
}
public static int TableauPositif(int tab[]) {
int taille = TailleNegatif(tab);
if(tab[taille] ==0)
taille+=1;
int taillepositif = TaillePositif(tab);
int tabPositif[] = new int[taillepositif];
for(int i = 0; i < tabPositif.length; i++) {
tabPositif[i] = tab[i + taille];
}
int min = tabPositif[0];
for(int i = 0; i< tabPositif.length; i++) {
if(min > tabPositif[i])
min = tabPositif[i];
}
return min;
}
public static int TailleNegatif(int tab[]) {
int cpt = 0;
for(int i = 0; i < tab.length; i++) {
if(tab[i] < 0) {
cpt +=1;
}
}
return cpt;
}
public static int TaillePositif(int tab[]) {
int cpt = 0;
for(int i = 0; i < tab.length; i++) {
if(tab[i] > 0) {
cpt +=1;
}
}
return cpt;
}
public static int distanceDeZero(int v1, int v2) {
int absv1 = v1 * (-1);
if(absv1 < v2)
return v1;
else if(absv1 > v2)
return v2;
else
return v2;
}
public static void main(String[] args) {
int t[] = {6,5,8,8,-2,-5,0,-3,-5,9,7,4};
Solution.trier(t);
}
}
To maintain O(n) time complexity and getting the desired results we have to add another variable called 'num' and assign to it 'near' before changing it's value. And finally make necessary checks. The improvements in the code are are:
public class CloseToZero {
public static void main(String[] args) {
int[] data = {2,3,-2};
int curr = 0;
int near = data[0];
int num=near;
// find the element nearest to zero
for ( int i=0; i < data.length; i++ ){
curr = data[i] * data[i];
if ( curr <= (near * near) ) {
num=near;
near = data[i];
}
}
if(near<0 && near*(-1)==num)
near=num;
System.out.println( near );
}
}
We have to find the Closest number to zero.
The given array can have negative values also.
So the easiest approach would append the '0' in the given array and sort it and return the element next to '0'
append the 0
Sort the Array
Return the element next to 0.
`
N = int(input())
arr = list(map(int, input().split()))
arr.append(0)
arr.sort()
zeroIndex = arr.index(0)
print(arr[zeroIndex + 1])
--> If this solution leaves corner cases please let me know also.
`
if you don't wanna use the inbuilt library function use the below code (just an and condition with your existing code)-
public class CloseToZero {
public static void main(String[] args) {
int[] data = {2,3,-2,-1,1};
int curr = 0;
int near = data[0];
// find the element nearest to zero
for ( int i=0; i < data.length; i++ ){
curr = data[i] * data[i];
if ( curr <= (near * near) && !((curr - (near * near) == 0) && data[i] < 0)) {
near = data[i];
}
}
System.out.println( near );
}
}
!((curr - (near * near) == 0) && data[i] < 0) : skip asignment if if near and curr is just opposit in sign and the curr is negative
public static int find(int[] ints) {
if (ints==null) return 0;
int min= ints[0]; //a random value initialisation
for (int k=0;k<ints.length;k++) {
// if a positive value is matched it is prioritized
if (ints[k]==Math.abs(min) || Math.abs(ints[k])<Math.abs(min))
min=ints[k];
}
return min;
}
public int check() {
int target = 0;
int[] myArray = { 40, 20, 100, 30, -1, 70, -10, 500 };
int result = myArray[0];
for (int i = 0; i < myArray.length; i++) {
if (myArray[i] == target) {
result = myArray[i];
return result;
}
if (myArray[i] > 0 && result >= (myArray[i] - target)) {
result = myArray[i];
}
}
return result;
}
I have added a check for the positive number itself.
Please share your views folks!!
public class ClosesttoZero {
static int closZero(int[] ints) {
int result=ints[0];
for(int i=1;i<ints.length;i++) {
if(Math.abs(result)>=Math.abs(ints[i])) {
result=Math.abs(ints[i]);
}
}
return result;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] ints= {1,1,5,8,4,-9,0,6,7,1};
int result=ClosesttoZero.closZero(ints);
System.out.println(result);
}
}
It can be done simply by making all numbers positive using absolute value then sort the Array:
int[] arr = {9, 1, 4, 5, 6, 7, -1, -2};
for (int i = 0; i < arr.length; ++i)
{
arr[i] = Math.abs(arr[i]);
}
Arrays.sort(arr);
System.out.println("Closest value to 0 = " + arr[0]);
import java.math.*;
class Solution {
static double closestToZero(double[] ts) {
if (ts.length == 0)
return 0;
double closestToZero = ts[0];
double absClosest = Math.abs(closestToZero);
for (int i = 0; i < ts.length; i++) {
double absValue = Math.abs(ts[i]);
if (absValue < absClosest || absValue == absClosest && ts[i] > 0) {
closestToZero = ts[i];
absClosest = absValue;
}
}
return closestToZero;
}
}
//My solution priorizing positive numbers contraint
int closestToZero = Integer.MAX_VALUE;//or we
for(int i = 0 ; i < arrayInt.length; i++) {
if (Math.abs(arrayInt[i]) < closestToZero
|| Math.abs(closestToZero) == Math.abs(arrayInt[i]) && arrayInt[i] > 0 ) {
closestToZero = arrayInt[i];
}
}
I need to write a java method sumAll() which takes any number of integers and returns their sum.
sumAll(1,2,3) returns 6
sumAll() returns 0
sumAll(20) returns 20
I don't know how to do this.
If your using Java8 you can use the IntStream:
int[] listOfNumbers = {5,4,13,7,7,8,9,10,5,92,11,3,4,2,1};
System.out.println(IntStream.of(listOfNumbers).sum());
Results: 181
Just 1 line of code which will sum the array.
You need:
public int sumAll(int...numbers){
int result = 0;
for(int i = 0 ; i < numbers.length; i++) {
result += numbers[i];
}
return result;
}
Then call the method and give it as many int values as you need:
int result = sumAll(1,4,6,3,5,393,4,5);//.....
System.out.println(result);
public int sumAll(int... nums) { //var-args to let the caller pass an arbitrary number of int
int sum = 0; //start with 0
for(int n : nums) { //this won't execute if no argument is passed
sum += n; // this will repeat for all the arguments
}
return sum; //return the sum
}
Use var args
public long sum(int... numbers){
if(numbers == null){ return 0L;}
long result = 0L;
for(int number: numbers){
result += number;
}
return result;
}
import java.util.Scanner;
public class SumAll {
public static void sumAll(int arr[]) {//initialize method return sum
int sum = 0;
for (int i = 0; i < arr.length; i++) {
sum += arr[i];
}
System.out.println("Sum is : " + sum);
}
public static void main(String[] args) {
int num;
Scanner input = new Scanner(System.in);//create scanner object
System.out.print("How many # you want to add : ");
num = input.nextInt();//return num from keyboard
int[] arr2 = new int[num];
for (int i = 0; i < arr2.length; i++) {
System.out.print("Enter Num" + (i + 1) + ": ");
arr2[i] = input.nextInt();
}
sumAll(arr2);
}
}
public static void main(String args[])
{
System.out.println(SumofAll(12,13,14,15));//Insert your number here.
{
public static int SumofAll(int...sum)//Call this method in main method.
int total=0;//Declare a variable which will hold the total value.
for(int x:sum)
{
total+=sum;
}
return total;//And return the total variable.
}
}
You could do, assuming you have an array with value and array length: arrayVal[i], arrayLength:
int sum = 0;
for (int i = 0; i < arrayLength; i++) {
sum += arrayVal[i];
}
System.out.println("the sum is" + sum);
I hope this helps.