allcoding1

INFOSYS ALL EXAM ANS ARE AVAILABLE FREE OFF COST @Infosys_Ans @Infosys_Ans @Infosys_Ans @Infosys_Ans NOT:-. ONCE CHECK IT

You are given a rooted tree of N vertices and an array A of N integers A[i] is the parent of the vertex (i+1 or 0 if the vertex (i+1) is the root. For each vertex V from 1 to N, the answer K is the total number of subsets of vertices with the LCA (lowest common ancestor) equal to V Since the of K can be large calculate it modulo 10 ^ 9 + 7 Let there be an integer array Res where Res[i] contains the answer for (I + 1)th  vertex, Find the array Res Notes: • The lowest common ancestor (LCA) is defined for X nodes A1, A2,.... Ax as the lowest node in the tree that has all A1 A2..... Ax as descendants (where we allow a node to be a descendant of itself) Input Formate The first line contains an integer N denoting the number of elements in A Each line 1 of the N subsequent lines (where 0<=i<N) contains an integer describing All It is given that A[i] denotes the parent of the vertex (i+1) or 0 If the vertex (i+1) is the root //@allcoding1 const int N = 100005; const int MOD = 1e9 + 7; int  a[N], dp[N], res[N]; vector<int> g[N]; void dfs(int u) {     dp[u] = 1;     for (int v : g[u]) {         dfs(v);         dp[u] = 1ll * dp[u] * (dp[v] + 1) % MOD;     }     res[u] = (1ll * dp[u] * (1 << g[u].size()) - 1 + MOD) % MOD; } vector<int> functionName(int n,vector<int>A) {     g=A;     dfs(1);     return res; }    Language c++

Permutation code Python

INFOSYS EXAM ANS 3PM ALL Slots are available Telegram:- @allcoding1                      @allcoding1 discussion group:- @Infosys_exam_Ans Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE= Share with your friends

INFOSYS EXAM ANS 10AM ALL Slots are available Telegram:- @allcoding1                      @allcoding1 discussion group:- @Infosys_exam_Ans Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE= Share with your friends

INFOSYS EXAM ANS 10AM ALL Slots are available Telegram:- @allcoding1                      @allcoding1 discussion group:- @Infosys_exam_Ans Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE= Share with your friends

Java INFOSYS EXAM ANS 10AM ALL Slots are available Telegram:- @allcoding1                      @allcoding1 discussion group:- @Infosys_exam_Ans Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE= Share with your friends

// Infosys // N flowers on a Recatangular pana int ans = 100000000; void solve(vector<int> a, int n, int k, int index, int sum,            int maxsum) {     if (k == 1)     {         maxsum = max(maxsum, sum);         sum = 0;         for (int i = index; i < n; i++)         {             sum += a[i];         }         maxsum = max(maxsum, sum);         ans = min(ans, maxsum);         return;     }     sum = 0;     for (int i = index; i < n; i++)     {         sum += a[i];         maxsum = max(maxsum, sum);         solve(a, n, k - 1, i + 1, sum, maxsum);     } } int GetMaxBeauty(int N, int K, vector<int> A) {     solve(A, N, K, 0, 0, 0);     return ans; } C++ INFOSYS EXAM ANS 10AM ALL Slots are available Telegram:- @allcoding1                      @allcoding1 discussion group:- @Infosys_exam_Ans Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE= Share with your friends

def getLargestString(s, k):     frequency_array = [0] * 26     for i in range(len(s)):         frequency_array[ord(s[i]) -                         ord('a')] += 1     ans = ""     i = 25     while i >= 0:         if (frequency_array[i] > k):             temp = k             st = chr( i + ord('a'))                          while (temp > 0):                 ans += st                 temp -= 1                        frequency_array[i] -= k             j = i - 1                          while (frequency_array[j] <= 0 and                    j >= 0):                 j -= 1             if (frequency_array[j] > 0 and                 j >= 0):                 str1 = chr(j + ord( 'a'))                 ans += str1                 frequency_array[j] -= 1                          else:                 break         elif (frequency_array[i] > 0):             temp = frequency_array[i]             frequency_array[i] -= temp             st = chr(i + ord('a'))             while (temp > 0):                 ans += st                 temp -= 1         else:             i -= 1                  return ans           if name == "main":        S = input()     k = 3     print (getLargestString(S, k)) Python Bob code Telegram:- @allcoding1

Python Once check it out put INFOSYS EXAM ANS 10AM ALL Slots are available Telegram:- @allcoding1                      @allcoding1 discussion group:- @Infosys_exam_Ans Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE= Share with your friends

Python INFOSYS EXAM ANS 10AM ALL Slots are available Telegram:- @allcoding1_official                      @allcoding1_official discussion group:- @Infosys_examAns Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE= Share with your friends

Java INFOSYS EXAM ANS 10AM ALL Slots are available Telegram:- @allcoding1_official                      @allcoding1_official discussion group:- @Infosys_examAns Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE= Share with your friends

def next_stepping_number(N): N = str(N) for i in range(len(N) - 1): if abs(int(N[i]) - int(N[i+1])) != 1: return int(N[:i+1] + str(int(N[i]) + 1 if int(N[i]) < int(N[i+1]) else int(N[i]) - 1) + '9'*(len(N)-i-1)) return int(N) + 1 print(next_stepping_number(4)) # should return 5 INFOSYS EXAM ANS 10AM ALL Slots are available Telegram:- @allcoding1 @allcoding1 discussion group:- @Infosys_exam_Ans Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE= Share with your friends

Python INFOSYS EXAM ANS 10AM ALL Slots are available Telegram:- @allcoding1                      @allcoding1 Discussion group:- @infosys_exam_Ans Coding Ans:- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE= Share with your friends

def shortest_subarray(colors, C):     n = len(colors)     color_count = [0] * (C+1)     left, right = 0, 0     min_length = float('inf')     count = 0     while right < n:         color_count[colors[right]] += 1         if color_count[colors[right]] == 1:             count += 1         while count == C:             min_length = min(min_length, right - left + 1)             color_count[colors[left]] -= 1             if color_count[colors[left]] == 0:                 count -= 1             left += 1         right += 1     if min_length == float('inf'):         return -1     else:         return min_length Python3