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last index of 1.cpp
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last index of 1.cpp
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/*
///////////////////////////////////////////
//Question/Info
Last index of One
Basic Accuracy: 67.57% Submissions: 4998 Points: 1
Given a string S consisting only '0's and '1's, find the last index of the '1' present in it.
Example 1:
Input:
S = 00001
Output:
4
Explanation:
Last index of 1 in given string is 4.
Example 2:
Input:
0
Output:
-1
Explanation:
Since, 1 is not present, so output is -1.
Your Task:
You don't need to read input or print anything. Your task is to complete the function lastIndex() which takes the string S as inputs and returns the last index of '1'. If '1' is not present, return "-1" (without quotes).
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(1)
Constraints:
1 <= |S| <= 106
S = {0,1}
Company Tags
InMobi
author: srj_v
///////////////////////////////////////////
*/
#include <bits/stdc++.h>
using namespace std;
// #define int long long int
#define sbit(x) __builtin_popcount(x)
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define eb(x) emplace_back(x)
#define ct(x) cout << x << "\n";
#define ct2(x,y) cout << x << " " << y << "\n";
#define tc(x) cout << x << " ";
#define tc2(x,y) cout << x << " " << y << " ";
#define forn(i,n) for(int i = 0; i < (int)(n); ++i)
#define forx(i,x,n) for(int i = x; i < (int)(n); ++i)
#define nfor(i,n) for(int i = n-1; i >= 0; --i)
#define all(v) v.begin(),v.end()
#define fsp(x,y) fixed << setprecision(y) << x
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007 // (1e9+7)
#define pii pair<int,int>
#define pis pair<int,string>
#define vi vector<int>
#define vii vector<pii>
#define mii map<int,int>
#define p_q priority_queue // priority_queue<int> (&) priority_queue< int,vi,greater<int> >
#define _IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long double ld;
typedef long long int lli;
#pragma GCC optimize("Ofast")
void c_p_c()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
}
int32_t main() {
///////////
c_p_c();
///////////
_IOS
//////////
// code
/*
int t ; cin >> t; while(t--){}
*/
class Solution {
public:
int lastIndex(string s)
{
for (int i = s.length() - 1; i >= 0; i--) {
if (s[i] == '1') {
return i;
}
}
return -1;
}
};
// cerr << "time: " << clock() << " ms" << '\n';
return 0;
}