HackerRank Yet Another KMP problem solution

In this HackerRank Yet Another KMP problem solution we have given a sequence construct a string S. if there are multiple strings that fulfill the conditions print the lexicographically smallest one.

Problem solution in Python.

import string
xs = list(map(int,input().split()))
ys = map(list,filter(lambda p: p[0] != 0, zip(xs, string.ascii_lowercase)))
ys = list(sorted(ys))
c = ys[0][1]
ys[0][0] -= 1
if ys[0][0] == 0:
    del ys[0]
ys = list(sorted(ys, key=lambda p: p[1]))
s = [c]
while ys:
    i = 0
    if len(s) >= 2 and len(ys) >= 2 and s[0] == s[1] == s[-1] == c == ys[i][1]:
        i = 1
    s.append(ys[i][1])
    ys[i][0] -= 1
    if ys[i][0] == 0:
        del ys[i]
print(*s, sep='')

Problem solution in Java.

import java.io.*;
import java.util.*;


public class KMP {
    public static void main(String[] args) {
        char ch='a',ch1;
        String s1="";
        int ar[]=new int [26];
        int min=99999999,loc=0;
        int loc2=0;
        Scanner in=new Scanner(System.in);
        int k=0;
        for(int i=0;i<26;i++){
            ar[i]=in.nextInt();
            if(ar[i]<min && ar[i]!=0){
               min=ar[i];loc=i;
            }
            if(ar[i]!=0){
                k++;
                if(k==2) {
                    loc2 = i;
                }
            }
        }
        ch1=(char)(97+loc);
        ar[loc]=ar[loc]-1;
        s1=s1+Character.toString(ch1);
        if(ar[loc]<ar[loc2]) {
            ch1 = (char) (97 + loc);
            char ch2 = (char) (97 + loc2);
            int len1 = ar[loc];
            if(ch1<ch2) {
                s1 = s1 + new String(new char[len1]).replace("", Character.toString(ch1) + Character.toString(ch2));
                ar[loc2] = ar[loc2] - len1;
                ar[loc] = ar[loc] - len1;
            }
        }

        for(int i=0;i<26;i++){
            String s=new String(new char[ar[i]]).replace("",Character.toString(ch));
            s1=s1+s;
            ++ch;
        }
        System.out.println(s1);
    }
}

Problem solution in C++.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

typedef unsigned int uint;
typedef unsigned long long uint64;
typedef long long sint64;

uint v[26];

char s[1000005];
uint sn;

int main(int argc, char* argv[])
{
	uint n = 26;

	for (uint i = 0; i < n; ++i)
		cin >> v[i];

	uint mi = 0;
	for (uint i = 0; i < 26; ++i)
	{
		if (v[i])
		{
			mi = i;
			break;
		}
	}

	uint mmi = mi;
	for (uint i = mi + 1; i < 26; ++i)
	{
		if (v[i] && v[i] < v[mmi])
			mmi = i;
	}

	s[sn++] = mmi + 'a';
	--v[mmi];
	if (mmi == mi)
	{
		for (uint i = mi + 1; i < 26; ++i)
		{
			if (v[i])
			{
				mmi = i;
				break;
			}
		}

		if (mi != mmi)
		{
			for (uint i = 0; i < v[mi]; ++i)
			{
				s[sn++] = mi + 'a';
				s[sn++] = mmi + 'a';
				--v[mmi];
			}
			v[mi] = 0;
		}
	}

	for (uint j = 0; j < 26; ++j)
	{
		for (uint i = 0; i < v[j]; ++i)
			s[sn++] = j + 'a';
	}

	cout << s << endl;
	return 0;
}

Problem solution in C.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int letter = 0, min=26, first = -1;
    int data[27];
    data[26]=1000001;
    for(int i = 0; i < 26; i++){
        scanf("%d",data+i);
        if(data[i]) {
            if (first<0) first = i;
            letter++;
            if(data[i]<data[min]) min = i;
        }
    }
    if(letter == 1) {
        for(int i = 0; i < data[min]; i++) {
            putchar('a'+min);
        }
        return 0;
    }
    if (min==first) {
        putchar('a'+min);
        int index_m = 1;
        for (int l = first + 1; l < 26; l++) {
            for (int i = 0; i<data[l]; i++) {
                if(index_m++ < data[min]) putchar('a'+min);
                putchar('a'+l);
            }
        }
    } else {
        putchar('a'+min);
        data[min]--;
        for (int l = first; l < 26; l++) {
            for (int i = 0; i<data[l]; i++) {
                putchar('a'+l);
            }
        }
    }
    
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}