Python program to solve the Exchange digits problem
Exchange digits problem
This Program was asked in TCS CodeVita where we need to print the next greatest number from the combinations of the other number. Here is the Python program to solve the Exchange digits problem by finding all the combinations of numbers to get the next greatest number. The level of this problem is good and can solve the problem using an inbuilt function permutation.
Problem Description
Compute the nearest larger number by interchanging its digits updated.Given 2 numbers a and b find the smallest number greater than b by interchanging the digits of a and if not possible print -1.
- Input Format
2 numbers a and b, separated by space.
- Output Format
A single number greater than b.If not possible, print -1
- Constraints
1 <= a,b <= 10000000
Example 1:
Sample Input:
459 500
Sample Output:
549
Example 2:
Sample Input:
645757 457765
Sample Output:
465577
Algorithm
- Step 1:- Start.
- Step 2:- From itertools import permutation function.
- Step 3:- Take user inputs.
- Step 4:- Initialize a flag variable to 0[zero].
- Step 5:- Convert number1 to String and then to a list.
- Step 6:- Sort the list by using the sorted function.
- Step 7:- Initialize a perm variable that stores the permutations of the list.
- Step 8:- Iterate through the perm variable by converting it to the list.
- Step 9:- Initialize an empty string.
- Step 10:- Concatenate all the iterators to a string variable.
- Step 11:- Typecast the string variable to an integer to check the next greatest number.
- Step 12:- Continue the process until if condition gets the True value.
- Step 13:- As the if condition gets the true value, chang the flag variable to 1 and break the loop.
- Step 14:- Check the flag whether we got the greatest value or not.
- Step 15:- Print the string variable if flag is 1 else print -1.
- Step 16:- End.
Python program to solve the Exchange digits problem
#import itertools to get permutation function from itertools import permutations #take inputs num1 = int(input('Enter the 1st number :')) num2 = int(input('Enter the 2nd number :')) #initialize a flag variable flag = 0 #convert num1 to string list num1 = list(str(num1)) #sort the list num1 = sorted(num1) #find all permutations perm = permutations(num1) #iterate through all permutations for i in list(perm): #initialize an string string = " " #iterate through an string for j in i: string+=j #typecast string to integer #check for next greater value if int(string) > num2: #if True Change the flag variable #break the loop flag = 1 break #check if the number is found or not if flag == 1: print(string) else: print(-1)
Output: Enter the 1st number :459 Enter the 2nd number :500 549
Staircase Problem in Other Coding Languages
C
Please contribute your code for For Solution of Exchange digit problem in C programing we will post it soon on the web site
C++
Please contribute your code for For Solution of Exchange digit problem in C++ programing we will post it soon on the web site
Java
Please contribute your code for For Solution of Exchange digit problem in Java programing we will post it soon on the web site
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## Solution using java
public class Solution {
static int res;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int a = sc.nextInt();
int b = sc.nextInt();
res = Integer.MAX_VALUE;
List list = new ArrayList();
while (a > 0) {
list.add(a % 10);
a /= 10;
}
backtrack(0, b, list, 0, new boolean[list.size()]);
System.out.println(res == b ? -1 : res);
}
private static void backtrack(int count, int b, List list, int num, boolean[] visited) {
if (count == list.size()) {
if (num > b) {
res = Math.min(res, num);
}
return;
}
for (int i = 0; i < list.size(); i++) {
if (!visited[i]) {
visited[i] = true;
backtrack(count + 1, b, list, num * 10 + list.get(i), visited);
visited[i] = false;
}
}
}
}
#without using permutation
def arrange_digits(n,t):
d_list = list(map(int,list(str(n))))
d_list.sort()
output = “”
for i in range(len(str(t))-1):
has_found = False
for j in range(len(d_list)):
if str(d_list[j])==str(t)[i]:
remain = “”
temp_arr = d_list.copy()
temp_arr.pop(j)
for r in temp_arr[::-1]:
remain += str(r)
if remain>str(t)[i+1::1]:
output += str(d_list.pop(j))
has_found = True
break
else:
continue
elif d_list[j]>int(str(t)[i]):
output += str(d_list.pop(j))
for d in d_list[::1]:
output += str(d)
return int(output)
if has_found:
has_found = False
else:
return -1
return int(output)
number,target = map(int,input().split())
print(arrange_digits(number,target))
from itertools import permutations
m = input()
n = input()
lst = sorted([int(“”.join(char)) for char in permutations(m)])
for i in lst:
if i > int(n):
print(i)
break
from itertools import permutations
a1,a2=input().split()
a=permutations(a1)
list1=[]
for i in a:
list1.append(int(“”.join(i)))
min1=1000000000
for i in range(len(list1)):
if(list1[i]>int(a2)):
if(list1[i]<min1):
min1=list1[i]
if(min1==1000000000):
print("-1")
else:
print(min1)
Thanks for contributing the code Vicky
x,y=input().split()
from itertools import permutations as pm
a=list(pm(x))
l=sorted(list(map(”.join,a)))
for i in l:
if i>y:
break
print(i)
thanks for contributing this code
C++ Solution
#include
using namespace std;
int fac(int n)
{
if(n==0 || n==1)
return 1;
else
return n*fac(n-1);
}
int main()
{
int a,b;
string s1,s2;
cin>>s1>>s2;
if(s1.size()>s2.size())
{
sort(s1.begin(),s1.end());
cout<<s1;
}
else if(s1.size()<s2.size())
{
cout<<"-1";
}
else
{
int i,j,k;
vector A;
for(i=0;i<fac(s1.size());i++)
{
next_permutation(s1.begin(),s1.end());;
A.push_back(s1);
}
sort(A.begin(),A.end());
stringstream ss1;
int b;
ss1<>b;
for(i=0;i<A.size();i++)
{
stringstream ss;
ss<>s;
if(s>b)
{
cout<<A[i];
break;
}
}
}
}
Thanks for contributing this code
n=int(input())
m=int(input())
grt_ele=[]
n=list(str(n))
perm=list(permutations(n))
res = list(map(“”.join, perm))
for i in res:
if int(m)<int(i):
grt_ele.append(i)
print(min(grt_ele))
Thanks for contributing the code Shivam