We Keep Coding

Each substring of a certian length of a binary substring should have at least one '1' character

You have been given a binary string containing only the characters '1' and '0'.
Calculate how many characters of the string need to be changed in order to make the binary string such that each of its substrings of at least a certain length contains at least one "1" character.
I came to think of the following idea but it fails for many testcases:
public static int minimumMoves(String s, int d) {
int n = s.length();
int i=0, answer = 0;
while(i<n)
{
boolean hasOne = false;
int j=i;
while(j<n && j<i+d)
{
if(s.charAt(j) == '1')
{
hasOne = true;
break;
}
j++;
}
if(!hasOne) {
answer++;
i += d;
}
else i++;
}
return answer;
}
Also my algorithm runs on O(|s|2) time. Can anyone suggest ideas on O(|s|) time?

Just throwing off an idea:
return s.split("(?<=\\G.{" + String.valueof(d) + "})").stream().filter(str -> str.contains("1")).count()

You just need to break ensure there is no run of d zeros.
public static int minimumMoves(String s, int d) {
int result = 0;
int runLength = 0;
for(char c: s.toCharArray()) {
if (c == '0') {
runLength += 1;
if (runLength == d) { // we need to break this run
result += 1;
runLength = 0;
}
} else {
runLength = 0;
}
}
return result;
}

I used the sliding window technique and Deque to solve this. This is my accepted solution:
public static int minimumMoves(String s, int d) {
int n = s.length();
Deque<Character> dq = new LinkedList<>();
int count = 0, answer = 0;
for(int i=0; i<d; i++)
{
if(s.charAt(i) == '1') count++;
dq.addLast(s.charAt(i));
}
if(count == 0) {
answer++;
count++;
dq.removeLast();
dq.addLast('1');
}
int i=d;
while(i<n)
{
if(dq.getFirst() == '1') count--;
dq.removeFirst();
if(s.charAt(i) == '1') count++;
dq.addLast(s.charAt(i));
if(count == 0)
{
answer++;
dq.removeLast();
dq.addLast('1');
count++;
}
i++;
}
return answer;
}

You just need to use a sliding window and a count of 1s so far at each index. Use a sliding window of d and if you don't see any ones so far, update the last index of that window with 1 and increment the result.
Code below:
public static int minimumMoves(String s, int d) {
int n = s.length();
int[] count = new int[n+1];
int res = 0;
for ( int i = 1; i <= d; i++ ) {
if ( s.charAt(i-1) == '1') count[i] = count[i-1]+1;
else count[i] = count[i-1];
}
if ( count[d] == 0 ) {
res++;
count[d] = 1;
}
for ( int i = d+1; i <= n; i++ ) {
if ( s.charAt(i-1) == '0' ) {
count[i] = count[i-1];
int ones = count[i] - count[i-d];
if ( ones == 0 ) {
count[i] = count[i-1] + 1;
res++;
}
} else {
count[i] = count[i-1] + 1;
}
}
return res;
}

Thought of another implementation you can do for this by working from the maximum possible changes (assumes at start that all values are '0' in String), reduce it when it finds a '1' value, and then jump to the next substring start. This allows it to run in O(n) and Ω(n/m) (n = String length, m = Substring length).
public static int minimumMoves(String s, int d)
{
char[] a = s.toCharArray();
//Total possible changes (not counting tail)
if(a.length < d)
return 0;
int t = (int) a.length / d;
//Total possible changes (counting tail)
//int t = (int) Math.ceil((double) a.length / (double) d);
for(int i = 0; i < a.length; i++)
{
if(a[i] == '1')
{
t--;
//Calculate index for start of next substring
i = (i / d + 1) * d - 1;
}
}
return t;
}

Traveling salesman code not working (Java)

Traveling salesman code in java below (gives wrong result)
http://www.sanfoundry.com/java-program-solve-travelling-salesman-problem-unweighted-graph/
package com.hinguapps.graph;
import java.util.InputMismatchException;
import java.util.Scanner;
public class TSP {
private int numberOfNodes;
private Stack < Integer > stack;
public TSP() {
stack = new Stack < Integer > ();
}
public void tsp(int adjacencyMatrix[][]) {
numberOfNodes = adjacencyMatrix[1].length - 1;
int[] visited = new int[numberOfNodes + 1];
visited[1] = 1;
stack.push(1);
int element, dst = 0, i;
int min = Integer.MAX_VALUE;
boolean minFlag = false;
System.out.print(1 + "\t");
while (!stack.isEmpty()) {
element = stack.peek();
i = 1;
min = Integer.MAX_VALUE;
while (i <= numberOfNodes) {
if (adjacencyMatrix[element][i] > 1 && visited[i] == 0) {
if (min > adjacencyMatrix[element][i]) {
min = adjacencyMatrix[element][i];
dst = i;
minFlag = true;
}
}
i++;
}
if (minFlag) {
visited[dst] = 1;
stack.push(dst);
System.out.print(dst + "\t");
minFlag = false;
continue;
}
stack.pop();
}
}
public static void main(String...arg) {
int number_of_nodes;
Scanner scanner = null;
try {
System.out.println("Enter the number of nodes in the graph");
scanner = new Scanner(System.in);
number_of_nodes = scanner.nextInt();
int adjacency_matrix[][] = new int[number_of_nodes + 1][number_of_nodes + 1];
System.out.println("Enter the adjacency matrix");
for (int i = 1; i <= number_of_nodes; i++) {
for (int j = 1; j <= number_of_nodes; j++) {
adjacency_matrix[i][j] = scanner.nextInt();
}
}
for (int i = 1; i <= number_of_nodes; i++) {
for (int j = 1; j <= number_of_nodes; j++) {
if (adjacency_matrix[i][j] == 1 &&
adjacency_matrix[j][i] == 0) {
adjacency_matrix[j][i] = 1;
}
}
}
System.out.println("The cities are visited as follows: ");
TSP tspNearestNeighbour = new TSP();
tspNearestNeighbour.tsp(adjacency_matrix);
} catch (InputMismatchException inputMismatch) {
System.out.println("Wrong Input format");
}
scanner.close();
}
}
Matrix should be :
0 10 5 40
2 0 5 1
6 13 0 12
1 8 9 0
Expected result: 1 3 2 4 1
Code result : 1 3 4 2 1

This implementation is wrong. This is a hard problem, because you need to either touch every path, or at the very least CONSIDER every path. This implementation basically boils down to "Each step, move to the closest node that I haven't visited". Since the stack is not keeping memory of where you have been, it does not backtrack to consider that a better path may have existed down one of the longer roads.
To fix this, the algorithm needs to keep the path in memory somehow, and not start printing the solution until the best solution has actually been found. (Can use recursion, a stack that holds the whole path, or some other method.)

Need help to solve the hackerrank challenge

I'm trying to solve an "Almost Sorted" challenge in hackerrank the problem is:
Given an array with elements, can you sort this array in ascending order using only one of the following operations?
Swap two elements.
Reverse one sub-segment.
Input Format
The first line contains a single integer, , which indicates the size of the array.
The next line contains integers separated by spaces.
Sample Input #1
2
4 2
Sample Output #1
yes
swap 1 2
Sample Input #2
3
3 1 2
Sample Output #2
no
Sample Input #3
6
1 5 4 3 2 6
Sample Output #3
yes
reverse 2 5
I tried to solve the challenge and my code is working but it seems it's to slow for big arrays.
Kindly asking you to help me to find a better solution for mentioned problem.
Below is my code:
import java.util.*;
public class Solution
{
private static int[] arr;
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
int N = in.nextInt();
arr = new int[N];
for (int i = 0; i < N; i++)
{
arr[i] = in.nextInt();
}
if (IsSorted(arr))
{
System.out.println("yes");
return;
}
if(CheckSingleSwap(arr))
return;
if(CheckSingleReverse(arr))
return;
System.out.println("no");
}
private static boolean CheckSingleReverse(int[] arr)
{
int length = arr.length;
int limit = length - 2;
int current = 1;
List<Integer> indexes = new ArrayList<Integer>();
while (current < limit)
{
for (int i = 0; i < length; i++)
{
int temp = current + i;
for (int j = i; j <= temp && temp < length; j++)
{
indexes.add(j);
}
if (IsSorted(ReverseArrayPart(arr, indexes)))
{
System.out.println("yes");
System.out.println("reverse " + (indexes.get(0) + 1) + " " + (indexes.get(indexes.size() - 1) + 1));
return true;
}
indexes.clear();
}
current++;
}
return false;
}
private static int[] ReverseArrayPart(int[] arr, List<Integer> indexes)
{
int[] result = new int[arr.length];
int[] arrayPart = new int[indexes.size()];
int j = 0;
for (int i = 0; i < arr.length; i++)
{
if (indexes.contains(i))
{
arrayPart[j] = arr[i];
j++;
}
result[i] = arr[i];
}
for(int i = 0; i < arrayPart.length / 2; i++)
{
int temp = arrayPart[i];
arrayPart[i] = arrayPart[arrayPart.length - i - 1];
arrayPart[arrayPart.length - i - 1] = temp;
}
j = 0;
for (int i = 0; i < result.length; i++)
{
if (indexes.contains(i))
{
result[i] = arrayPart[j];
j++;
}
}
return result;
}
private static boolean CheckSingleSwap(int[] arr)
{
int count = 0;
int[] B = Arrays.copyOf(arr, arr.length);
Arrays.sort(B);
List<Integer> indexes = new ArrayList<Integer>();
for(int i = 0; i < arr.length; i++)
{
if(arr[i] != B[i])
{
count++;
indexes.add(i+1);
}
}
if(count > 2)
return false;
System.out.println("yes");
System.out.println("swap " + indexes.get(0) + " " + indexes.get(1));
return true;
}
private static boolean IsSorted(int[] arr)
{
int length = arr.length;
for (int i = 0; i < length - 1; i++)
{
if (arr[i] > arr[i + 1])
{
return false;
}
}
return true;
}
}

For the following code, pass in A as the original array and B as the sorted array.
CheckSingleSwap:
Instead of adding the indices to another list, store the first swap you encounter, and keep going; if you find the corresponding other swap, then store it and record the finding; if you find a different swap, exit with false. At the end if you've recorded the finding, print the corresponding indices.
private static boolean CheckSingleSwap(int[] A, int[] B)
{
int L = A.length;
int firstSwap = -1, secondSwap = -1;
for(int i = 0; i < L; i++)
{
if(A[i] != B[i])
{
if (firstSwap == -1)
firstSwap = i;
else if (secondSwap == -1 && A[i] == B[firstSwap] && A[firstSwap] == B[i])
secondSwap = i;
else
return false;
}
}
if (firstSwap != -1 && secondSwap != -1)
{
System.out.println("yes");
System.out.println("swap " + (firstSwap + 1) + " " + (secondSwap + 1));
return true;
}
System.out.println("array is already sorted!");
return false; // or whatever you decide to do; maybe even an exception or enumerated type
}
CheckSingleReverse:
You are doing WAY too much here! You seem to be brute forcing every single possible case (at first glance).
What you can do instead is to find the region where all the numbers are different. If there are more than two of these, or two which are separated by more than one element, then return false immediately.
The reason for the "more than one" thing above is because of odd-number length regions - the middle element would be the same. If you find such two regions, treat them as one. Then you can proceed to find out if the region is reversed.
private static boolean CheckSingleReverse(int[] A, int[] B)
{
// find region
int L = A.length;
int diffStart = -1, diffEnd = -1; boolean mid = false, found = false;
for (int i = 0; i < L; i++)
{
if (A[i] != B[i])
{
if (found)
{
if (i - diffEnd == 2 && !mid)
{
mid = true;
found = false;
diffEnd = -1;
}
else
return false;
}
else if (diffStart == -1)
diffStart = i;
}
else
if (diffStart != -1 && diffEnd == -1)
{
found = true;
diffEnd = i - 1;
}
}
if (diffEnd == -1)
{
if (A[L - 1] != B[L - 1])
diffEnd = L - 1;
else if (!found)
{
System.out.println("array is already sorted!");
return false;
}
}
// find out if it's reversed
int count = (diffEnd - diffStart + 1) / 2;
for (int i = 0; i < count; i++)
{
int oneEnd = diffStart + i, otherEnd = diffEnd - i;
if (!(A[oneEnd] == B[otherEnd] && A[otherEnd] == B[oneEnd]))
return false;
}
System.out.println("yes");
System.out.println("reverse " + (diffStart + 1) + " " + (diffEnd + 1));
return true;
}
Just to give you an idea of the performance boost, on ideone.com, with an array length of 150, the original implementation of CheckSingleReverse took 1.83 seconds, whereas the new one took just 0.1 seconds. With a length of 250, the original actually exceeded the computational time limit (5 seconds), whereas the new one still took just 0.12 seconds.
From this it would seem that your implementation takes exponential time, whereas mine is linear time (ignoring the sorting).
Funnily enough, with an array size of 3 million I'm still getting around 0.26 seconds (ideone's execution time fluctuates a bit as well, probs due to demand)

Zigzag conversion

Question is : The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N
A P L S I I G
Y I R
And then read line by line: "PAHNAPLSIIGYIR"
I have written below code, appearantly it works fine, but I might miss some corner cases. Could you help me to find all the corner cases for this question on my answer?
public static String zigZagConversion(String s , int rowNum){
if (s == null){
throw new IllegalArgumentException();
}
if (rowNum == 1){
return s;
}
StringBuilder str = new StringBuilder();
int step = 2 * rowNum - 2 ;
for (int i = 0 ; i < rowNum ; i++){
if( i == 0 || i == rowNum -1){
for (int j = i ; j < s.length() ; j +=step){
str.append(s.charAt(j));
}
}
else{
int step2 = 2* (rowNum - i - 1);
int step3 = step - step2;
int k = i;
boolean flag = true;
while (k < s.length()){
str.append(s.charAt(k));
if(flag){
k += step2;
flag = false;
}
else{
k +=step3;
flag = false;
}
}
}
}
return str.toString();
}

It gives incorrect output for "PAYPALISHIRING", 4
P I N
A L S I G
Y A H R
P I
So the correct answer should be PINALSIGYAHRPI.
But your program gives PINALIGYAIHRNPI:
an "S" is missing, one extra "I" and one extra "N".
Your revised version is still incorrect, it gives PINALSIIGYAHNPI.
The problem is in the while loop in the middle.
You need to alternate the step counting,
setting the flag on and off.
Your mistake was to only set it off once, and never back on again.
str.append(s.charAt(k));
if (flag) {
k += step2;
flag = false;
} else {
k += step3;
flag = true;
}
With this correction, I believe your solution is correct. (I also added a minor improvement there, extracting the common str.append(s.charAt(k)); from the if-else branches.

My solution on leetcode forum:
https://leetcode.com/problems/zigzag-conversion/discuss/549451/Java-Solution-O(n)-with-algorithm
The mathematic algorithm for zigzag is:
originalDiff = numRows * 2 - 2;
If -> 'currRow' equals First or last lines
use the originalDiff (numRows * 2 - 2)
Else ->
For each new line:
upperDiff += 2,
lowerDiff -=2
Examples:
numRows =2 -> originalDiff = 2
PYAIHRN
APLSIIG
3 -> 4
P A H N
A P L S I I G
Y I R
numRows = 4 -> originalDiff = 6
P I N
A L S I G
Y A H R
P I
numRows = 5 -> originalDiff = 8
P H
A SI
Y I R
P L I G
A N
*/
My solution:
class Solution {
public String convert(String s, int numRows) {
if(numRows == 1) {
return s;
}
String newString = "";
int originalDiff = numRows * 2 - 2;
int diff = originalDiff;
int upperDiff = 0;
boolean isGoingDown = true;
int currIndex = 0;
int currRow = 0;
int startingIndex = 0;
for(int i = 0; i < s.length(); i++) {
System.out.println(currIndex);
newString += s.charAt(currIndex);
if(currRow == 0 || currRow == numRows - 1) {
currIndex += originalDiff;
} else {
if(isGoingDown) {
currIndex += diff;
isGoingDown = !isGoingDown;
} else {
currIndex += upperDiff;
isGoingDown = !isGoingDown;
}
}
if(currIndex >= s.length()) {
currRow++;
diff -= 2;
upperDiff += 2;
currIndex = currRow;
isGoingDown = true;
}
if(currRow == numRows) {
i = s.length();
}
}
return newString;
}
}

Zigzag conversion from leetcode in Javascript
Solution
const zigzag = (str, num) => {
if (num === 1) {
return str;
}
let check = true;
let result = [];
let i = 0;
while (i < str.length) {
result.push([]);
let j = 0;
while (j < num) {
if (check){
result[result.length-1].push(str[i]);
i++;
} else {
if (j == 0) {
result[result.length-1].push(null);
} else if (j === num-1) {
result[result.length-1].unshift(null);
} else {
result[result.length-1].unshift(str[i]);
i++;
}
}
j++;
}
check = !check;
}
let zigzag = [];
for (let k = 0; k < num; k++){
for(let l = 0; l < result.length; l++) {
zigzag.push(result[l][k]);
}
}
return zigzag.join("");
}
Example Input
zigzag("ABCD", 3)
Output
ABDC
Run
https://repl.it/#VinitKhandelwal/zigzag-conversion-javascript

Using HashMap
public String convert(String s, int numRows) {
if (numRows == 1){
return s;
}
StringBuilder result = new StringBuilder();
Map<Integer, StringBuilder> map = new HashMap<>();
for (int i = 0; i < numRows; i++) {
map.put(i,new StringBuilder());
}
int it = 0;
boolean flip = true;
for (int i = 0; i < s.length(); i++) {
if (flip) {
if(it<s.length()){
map.get(it).append(s.charAt(i));
it++;
}
} else {
map.get(it).append(s.charAt(i));
it--;
}
if (it + 1 == numRows || it == 0)
flip = !flip;
}
for (Map.Entry entry: map.entrySet()) {
result.append(entry.getValue());
}
return result.toString();
}

My Solution is traversing the string in the same way it is said in the problem, it is better to make string array of size numrows and the rest is storing the string character as it is in the logic,
you can keep the index and when that index is 0 i.e at the starting then we have to go till the end of the row and then except for first and last row, every array will have diagonal element.
So after traversing till the end then assign index = numrows - 2 and save in the respective array string and decrease and do the same till index >0 and then again traverse till the end row, do this and when we reach the end of the string then break from the loop.
and then concate all the string of string array in a new res string.
class Solution {
public String convert(String s, int n) {
if(n==1 || n>=s.length())
return s;
String[] a = new String[n]; //string array
int ind=0; // index for the string array
boolean flag=true;
int cnt=0; //to keep the counter till where we have traversed the string
while(true && flag)
{
if(ind==0)
{
for(int i=0;i<n;i++)
{
a[i] += s.charAt(cnt);
cnt++;
if(cnt==s.length())
{
flag=false;
break;
}
} // here it has reached the end so we assign here
ind = n-2;
}
else if(ind>0 && ind<n && flag)
{
a[ind] += s.charAt(cnt);
cnt++;
if(cnt==s.length())
{
flag=false;
break;
}
ind--; // to move diagonally up
}
}
String res = new String("");
for(int i=0;i<a.length;i++)
{
// System.out.println(a[i].substring(4));
res += a[i].substring(4);
}
return res;
}
}

Following is the simple solution.
class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows <= 1:
return s
res = ""
p = numRows * 2 - 2
temp = p
for i in range(0,numRows):
index = i
flag = 0
while index < len(s):
res = res + s[index]
if i == 0 or i == numRows-1:
index = index + p
else:
if flag == 0:
index = index + temp
flag = 1
else:
index = index + p-temp
flag = 0
temp = temp - 2
return res

zigzag-conversion Complete JavaScript-based solution
Created an Array of an array of row lengths. The main motive is to arrange characters in 2D array form and concat string row-wise.
var convert = function(s, numRows) {
let array =[],c=0,str='';
for(let row =0; row<numRows ; row++) {
array[row] = new Array();
}
while(c < s.length) {
for(let row =0; row<numRows ; row++) {
if((row+1)%numRows ==0) {
array[row].push(s[c]);
c++;
break;
} else {
array[row].push(s[c]);
c++;
}
}
for(let rr = numRows-2 ; rr>0;rr--) {
array[rr].push(s[c]);
c++;
}
}
for(let row =0; row<numRows ; row++) {
for(let i=0;i<array[row].length;i++){
if(array[row][i]){
str+=array[row][i]
}
}
}
return str
};
convert("PAYPALISHIRING",3)

Prime number generator between 2 nums

I cant figure out which test cases the code provided below fails.
problem:
All submissions for this problem are available.
Shridhar wants to generate some prime numbers for his cryptosystem. Help him!
Your task is to generate all prime numbers between two given numbers.
Input
The first line contains t, the number of test cases (less then or equal to 10).
Followed by t lines which contain two numbers m and n (1 <= m <= n <= 1000000000, n-m<=100000) separated by a space.
Output
For every test case print all prime numbers p such that m <= p <= n,
one number per line. Separate the answers for each test case by an empty line.
Example
Input:
2
1 10
3 5
Output:
2
3
5
7
3
5
import java.util.Scanner;
public class Main implements Runnable {
public static void main(String args[]) {
new Main().run();
}
#Override
public void run() {
Scanner sc = new Scanner(System.in);
Integer d;
try {
d = sc.nextInt();
boolean isPrime[] = new boolean[100000];
for (int i = 0; i < d; i++) {
int m = sc.nextInt();
int n = sc.nextInt();
if (n <= 0 || m > n) {
continue;
}
if (m <= 0) {
m = 2;
if (m > n) {
continue;
}
}
if (m == 1) {
m = 2;
}
if (m == 2 && n - m == 0) {
System.out.println(2);
} else {
for (int k = 0; k <= n - m; k++) {
isPrime[k] = true;
}
int sqrt = (int) Math.sqrt(n);
for (int j = 2; j <= sqrt; j++) {
int k = (m % j == 0) ? m / j : (m + j) / j;
for (; k <= n / j; k++) {
if (!(m == 2 && (j * k == 2)) && k != 1) {
isPrime[j * k - m] = false;
}
}
}
for (int a = m; a <= n; a++) {
if (isPrime[a - m]) {
System.out.println(a);
}
}
}
System.out.println();
}
} catch (Exception e) {
e.printStackTrace();
} finally {
sc.close();
}
}
}

Your problem is with this:
boolean isPrime[] = new boolean[100000];
and this:
for (int k = 0; k <= n - m; k++) {
isPrime[k] = true;
}
because some time n - m = 100000 then you need isPrime[100000] which you didnt allocate so you need to declare isPrime like this:
boolean isPrime[] = new boolean[100001];