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A permutation where each element indicates either number of elements before or after it?
In this section we will see one problem. Here n elements are given in an array. We have to check whether there is a permutation of that array exists, such that each element indicates the number of elements present either before or after it.
Suppose the array elements are {2, 1, 3, 3}. The appropriate permutation is like {3, 1, 2, 3}. Here the first 3 is indicating there are three elements next of it, the 1 indicates there is only one element before this. The 2 indicates there are two elements before it and the last 3 indicates that there are three elements before it.
Algorithm
checkPermutation(arr, n)
begin define a hashmap to hold frequencies. The key and value are of integer type of the map. for each element e in arr, do increase map[e] by 1 done for i := 0 to n-1, do if map[i] is non-zero, then decrease map[i] by 1 else if map[n-i-1] is non-zero, then decrease map[n-i-1] by 1 else return false end if done return true end
Example
#include<iostream>
#include<map>
using namespace std;
bool checkPermutation(int arr[], int n) {
map<int, int> freq_map;
for(int i = 0; i < n; i++){ //get the frequency of each number
freq_map[arr[i]]++;
}
for(int i = 0; i < n; i++){
if(freq_map[i]){ //count number of elements before current element
freq_map[i]--;
} else if(freq_map[n-i-1]){ //count number of elements after current element
freq_map[n-i-1]--;
} else {
return false;
}
}
return true;
}
main() {
int data[] = {3, 2, 3, 1};
int n = sizeof(data)/sizeof(data[0]);
if(checkPermutation(data, n)){
cout << "Permutation is present";
} else {
cout << "Permutation is not present";
}
}Output
Permutation is present
Updated on: 30-Jul-2019
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