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Additive Number in C++
Suppose we have a string containing only digits through '0' to '9', we have to write a function to determine whether it's an additive number or not. The additive number is a string whose digits can form additive sequence. A valid additive sequence should contain at least three numbers. Here except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two. So if the input is like “112358”, then the answer will be true, as 2 = 1 + 1, 3 = 1 + 2, 5 = 2 + 3, 8 = 3 + 5.
To solve this, we will follow these steps −
Define a method called ok(), this will take s, index, prev1, prev2
if index >= size of s, then return true
req := prev1 + prev2 and num := req as string
x := one blank string
for i in range index to size of s
x := x + s[i]
if x = num, and ok(s, i + 1, prev2, x as integer), then return true
return false
From the main method do the following −
n := size of num
for i in range 1 to n – 2
for j in range 1 to i
s1 := substring of num from 0 to j – 1
s2 := substring of num from j to i – j
x := max of s1 size and s2 size
if x > n – i, then go for the next iteration
if (s1[0] is 0 and size of s1 > 0) OR (s2[0] is 0 and size of s2 > 1), then skip to the next iteration
if ok(num, i + 1, s1 as integer and s2 as integer) is true, then return true
return false
Example(C++)
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
public:
bool ok(string s, int idx, lli prev1, lli prev2){
if(idx >= s.size()) return true;
lli req = prev1 + prev2;
string num = to_string(req);
string x = "";
for(int i = idx; i < s.size(); i++){
x += s[i];
if(x == num && ok(s, i + 1, prev2, stol(x))) return true;
}
return false;
}
bool isAdditiveNumber(string num) {
int n = num.size();
for(int i = 1; i < n - 1; i++){
for(int j = 1; j <= i; j++){
string s1 = num.substr(0, j);
string s2 = num.substr(j, i - j + 1);
int x = max((int)s1.size(), (int)s2.size());
if(x > n - i) continue;
if((s1[0] == '0' && s1.size() > 1) || (s2[0] == '0' && s2.size() > 1)) continue;
if(ok(num, i + 1, stol(s1), stol(s2))) return true;
}
}
return false;
}
};
main(){
Solution ob;
cout << (ob.isAdditiveNumber("112358"));
}Input
"112358"
Output
1
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