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Print an array with numbers having 1, 2 and 3 as a digit in ascending order
Here, the task is to print those number in an array having 1, 2 and 3 as digits in their numbers and if their is no such number than the output must be -1
Input : arr[] = {320,123,124,125,14532,126,340,123400,100032,13,32,3123,1100}
Output : 123 3123 14532 100032 123400
Since the array have values with digits 1, 2 and 3 it wouldn’t return -1 and print 5 values that
Contain 1, 2 and 3 in their respective numbers.Algorithm
START
Step 1 -> Declare array with elements of int type as arr
Step 2 -> store size of array in int n
Step 3 -> declare int variable as one, two, three
Step 4 -> call sort functions with parameters as arr and arr+n
Step 5 -> declare variable of type osrtingstream as st and string as num
Step 6 -> Loop For i=0 and i<n and ++i
Set one=two=three=1
Print arr[i]
Set num=st.str()
Set one=num.find("1")
Set two=num.find("2")
Set three=num.find("3")
IF((one!=-1)&&(two!=-1)&&(three!=-1))
Print num
End
Call st.str(‘’”)
end
STOPExample
#include <bits/stdc++.h>
#include<string.h>
#include<sstream>
using namespace std;
int main() {
int arr[] = {320,123,124,125,14532,126,340,123400,100032,13,32,3123,1100};
int n = sizeof(arr)/sizeof(arr[0]);
int one,two,three;
sort(arr, arr+n);
ostringstream st;
string num;
for (int i = 0; i < n; ++i) {
one=two=three=-1;
st << arr[i];
num=st.str();
one=num.find("1");
two=num.find("2");
three=num.find("3");
if((one!=-1)&&(two!=-1)&&(three!=-1)) {
cout<<num<<" ";
}
st.str("");
}
}Output
if we run the above program then it will generate the following output
123 3123 14532 100032 123400
Updated on: 30-Jul-2019
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