Find the string of length N according to the given conditions
Last Updated :
29 Jan, 2024
Given two integers N and K. Then your task is to output the string of length N such that the cumulative sum of the string is equal to K:
- Where the Cumulative sum of the string is: For each character add 2pos into the sum. Where pos is equivalent to the position of the character in the alphabet series. For example: a = 0, b = 1, c = 2 . . . . z = 25.
Note: If no such string is possible then output -1. If there are multiple answers, then print any of them.
Examples:
Input: N = 5, K = 13
Output: dbaaa
Explanation: The value of Z for d, b and a are 3, 1 and 0. Then cumulative sum of dbaaa is: (23) + (21) + (3*(20)) = 8 + 2 + 3 = 13. So both conditions met, length of output string is N = 5 and cumulative sum is K = 13.
Input: N = 4, K = 3
Output: -1
Explanation: It can be verified that no such string is possible.
Approach: Implement the idea to solve the problem
The problem is based on the Greedy logic and some observation. The problem can be divided into sub-parts as:
- If (N > K)
- No string of length N is possible in such cases. Because Minimum power of 2 is 1 so we need the cumulative sum to be at least equal to N.
- If (N == K)
- The answer will be a string of length N with all characters equal to ‘a’.
- If (N < K)
- Define an array let say power[], which is initialized to store the powers of 2.
- Find the highest power of 2 that is less than or equal to K and store it in a variable let say P.
- While there are still characters left to add (N > 0), calculate the remaining value of K after subtracting the current power of 2 (remk = K – power[P]). If this remaining value is less than the number of characters left to add minus one (remk < N – 1), it decreases P by 1. Otherwise, it adds a character corresponding to the current power of 2 to the string, decreases N by 1, and updates K to the remaining value.
- If there’s still a remaining value after all characters have been added (K > 0), it means a string couldn’t be formed, so -1 is returned, else return the answer string.
Steps were taken to solve the problem:
- Create an array let say Power[] of length 26.
- Run a loop and initialize 2’s power into Power[] from 20 to 225
- Create a StringBuilder let say Sb
- If (N > K), append “-1” to Sb
- Else if (N == K) append ‘a’ to Sb N times.
- Else if (N < K),
- Find the highest power of 2 that is less than or equal to K and store it in a variable, P.
- While (N > 0)
- Initialize remk with (K – Power[P])
- If(remk < (N – 1)), decrease P by 1
- Else, append character corresponding to P to sb, decrease N by 1 and update K = remK
- If (K > 0)
- Output Sb.
Code to implement the approach:
C++
#include <iostream>
#include <cmath>
#include <string>
using namespace std;
void solve(int N, int K) {
int power[26];
for (int i = 0; i < 26; i++) {
power[i] = pow(2, i);
}
string sb = "";
if (N > K) {
sb = "-1";
}
else if (N == K) {
for (int i = 0; i < N; i++) {
sb += 'a';
}
}
else {
int i = 0;
while (i < 26) {
if (power[i] == K || power[i] > K) {
break;
}
i++;
}
int P = i - 1;
while (N > 0) {
int remk = K - power[P];
if (remk < N - 1) {
P--;
}
else {
sb += (char)(97 + P);
N--;
K = remk;
}
}
if (K > 0) {
sb = "-1";
}
}
cout << sb << endl;
}
int main() {
int N = 5;
int K = 13;
solve(N, K);
return 0;
}
|
Java
import java.util.*;
public class GFG {
public static void main(String[] args)
{
int N = 5;
int K = 13;
solve(N, K);
}
public static void solve(int N, int K)
{
int power[] = new int[26];
for (int i = 0; i < power.length; i++) {
power[i] = (int)Math.pow(2, i);
}
StringBuilder sb = new StringBuilder();
if (N > K) {
sb.append("-1");
}
else if (N == K) {
for (int i = 0; i < N; i++) {
sb.append("a");
}
}
else {
int i = 0;
for (i = 0; i < power.length; i++) {
if (power[i] == K || power[i] > K) {
break;
}
}
int P = i - 1;
while (N > 0) {
int remk = K - power[P];
if (remk < N - 1) {
P--;
}
else {
sb.append((char)(97 + P));
N--;
K = remk;
}
}
if (K > 0) {
sb = new StringBuilder();
sb.append("-1");
}
}
System.out.println(sb.toString());
}
}
|
Python3
def solve(N, K):
power = [2**i for i in range(26)]
sb = []
if N > K:
sb.append("-1")
elif N == K:
sb.append("a" * N)
else:
i = 0
while i < len(power):
if power[i] == K or power[i] > K:
break
i += 1
P = i - 1
while N > 0:
remk = K - power[P]
if remk < N - 1:
P -= 1
else:
sb.append(chr(97 + P))
N -= 1
K = remk
if K > 0:
sb = ["-1"]
print("".join(sb))
def main():
N = 5
K = 13
solve(N, K)
if __name__ == "__main__":
main()
|
C#
using System;
using System.Text;
public class GFG
{
public static void Main(string[] args)
{
int N = 5;
int K = 13;
Solve(N, K);
}
public static void Solve(int N, int K)
{
int[] power = new int[26];
for (int i = 0; i < power.Length; i++)
{
power[i] = (int)Math.Pow(2, i);
}
StringBuilder sb = new StringBuilder();
if (N > K)
{
sb.Append("-1");
}
else if (N == K)
{
for (int i = 0; i < N; i++)
{
sb.Append("a");
}
}
else
{
int i = 0;
for (i = 0; i < power.Length; i++)
{
if (power[i] == K || power[i] > K)
{
break;
}
}
int P = i - 1;
while (N > 0)
{
int remk = K - power[P];
if (remk < N - 1)
{
P--;
}
else
{
sb.Append((char)(97 + P));
N--;
K = remk;
}
}
if (K > 0)
{
sb = new StringBuilder();
sb.Append("-1");
}
}
Console.WriteLine(sb.ToString());
}
}
|
Javascript
function solve(N, K) {
const power = [];
for (let i = 0; i < 26; i++) {
power[i] = Math.pow(2, i);
}
let sb = "";
if (N > K) {
sb = "-1";
} else if (N === K) {
for (let i = 0; i < N; i++) {
sb += 'a';
}
} else {
let i = 0;
while (i < 26) {
if (power[i] === K || power[i] > K) {
break;
}
i++;
}
let P = i - 1;
while (N > 0) {
const remk = K - power[P];
if (remk < N - 1) {
P--;
} else {
sb += String.fromCharCode(97 + P);
N--;
K = remk;
}
}
if (K > 0) {
sb = "-1";
}
}
console.log(sb);
}
const N = 5;
const K = 13;
solve(N, K);
|
Time Complexity: O(N), where N is the length of the output string
Auxiliary Space: O(26) ~ O(1)
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