Accenture Exact Questions Asked on 1st September | Accenture Actual Questions 

n = int(input()) c = 0 if n < 3: print(n) elif n in range(3, 13): print(n-1) elif n >= 13: for i in range(n): ele = str(i) if '3' in ele: c += 1 print(n-c) It is also helpful I hope

Problem 2 int n ; cin >> n; vector arr(n); for(auto & it : arr) cin >> it; int cnt = count(arr.begin(), arr.end() , 5); int rem = n - cnt; cnt /= 9; cnt *= 9; string ans = ""; while(cnt--) ans += '5'; while(rem--) ans += '0'; cout

In Python it's super easy: def count_not_3(n): count = 0 for i in range(1,n+1): if '3' in str(i): continue count += 1 return count

problem 1....simple solution (in java) :- public static int countNumbersWithout3(int n) { int count = 0; for (int num = 1; num 0) { int digit = current % 10; if (digit == 3) { hasThree = true; break; // No need to continue checking if we found a 3 } current /= 10; } if (!hasThree) { count++; } } return count; }

how we will be able to think of this solution in exam!!!! I mean i'm seeing this code for the very first time

int countNumbersWithThree(int num) { int count = 0; for (int i = 0; i < num; i++) { if (to_string(i).find('3') != string::npos) { count++; } } return count; }

n=input() count=0 for i in range(int(n)) : c=str(i) for j in range(len(c)) : if c[j] == '3' : count = count + 1 break a=int(n) print(a-count) will it be ok for time constraint mam

for count5 = total5 - (total5%9) will also work

Has anyone received actual assessment link after completing the screening test? I had given my screening test on September 1st.

Typical Digit DP problem 1 #include using namespace std; #define int long long int dp[100][100][2]; int solve(int ind, int cnt, int flag, string & s ) { if(ind == s.size()) { return cnt==0; } if(dp[ind][cnt][flag] != -1) return dp[ind][cnt][flag]; int ans = 0 ; if(!flag) { for(int i = 0 ; i < s[ind] - '0'; i++) { ans = ans + (solve(ind + 1, cnt + ( i == 3), 1, s)); } ans = ans + solve(ind + 1, cnt + (s[ind] == '3'), 0, s); } else { for(int i = 0 ; i < 10 ; i++) { ans = ans + solve(ind + 1, cnt + (i == 3), 1, s); } } return dp[ind][cnt][flag] = ans; } int32_t main() { ios_base::sync_with_stdio(false); cin.tie(NULL); memset(dp, -1, sizeof(dp)); int n ; cin >> n; string str =to_string(n); cout

in c++ #include #include using namespace std; int main() { int num; cin>>num; int count =0; int basecase =0; for(int i = 10; i

Can someone tell me are the ms office and network security sections mandatory for ase role?

mam do you have 2nd september 2023 questions

What are the languages that can be used ? Can python be used?

Mam is there any Time complexity constraint in accenture coding questions??? Like can we do this question in O(n2) time complexity if i didn't remember this solution??

Straight forward C++ sol ( not using string ) int count(int n) { int ans=0; if(n

JAVA ALSO HAVE PREDEFINED STRING FUNCTION : public class Main { public static void main(String[] args) { int n = 45; int result = countNot3(n); System.out.println(result); } public static int countNot3(int n) { int count = 0; for (int i = 1; i

Nice

Worst explanation of code