Code With Virus

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Today Morning time asked questions with solutions

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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 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++ Infosys @codewithvirus @codewithvirus @codewithvirus https://bit.ly/infosys-SP-DSP

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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 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++ Infosys @codewithvirus @codewithvirus @codewithvirus https://bit.ly/infosys-SP-DSP

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You are given Q queries, Each query contains a number N=Quer[i] denoting the ith query. You have to find M such that: 1 <= M <= N M|M+1|....| N is as maximum as possible where | is the OR bitwise operation. M is as maximum as possible. The answer to this query is the value of M. Find the sum of answers to all queries modulo 109+7 Note: A bitwise OR is a binary operation that takes two-bit patterns of equal length and performs the logical inclusive OR operation on each pair of corresponding bits. The result in each position is 0 if both bits are 0, while otherwise, the result is 1. For example, 0101 (decimal 5) OR 0011 (decimal 3) =0111(decimal 7 ) Input formate: The first line contains an integer, Q, denoting the number of elements in quer Each line i of the Q subsequent lines (where 0 <= 1 < Q ) contains an integer describing Quer[i] const int mod = 1e9 + 7; int f(int N) { return (1 << (bitset<32>(N).count() - 1)) - 1; } int solve(int Q,vector<int>Quer) { int ans = 0; for(int i=0;i<Q;i++) ans += f(Quer[i]); ans %= mod; } return ans; } Language c++ Infosys @codewithvirus @codewithvirus @codewithvirus https://bit.ly/infosys-SP-DSP

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you are given an array A of N elements and an integer K, A subsequence is good if it is a non-decreasing subsequence and the sum of its elements is at least k. Find the minimum length of a good subsequence, or return-1 if there is no such subsequence Note: A subsequence of array A is a sequence that can be derived from array A by deleting some or no elements without changing the order of the remaining elements. For example, [2, 4, 6] is a subsequence of [1,2,3,4,5, 6, 7] but [3, 4, 1] is not. A subsequence is non-decreasing every element in the subsequence is greater or equal to its previous element. The length of the subsequence is the number of elements in it. Input Format: The first line contains an integer N. denoting the number of elements in A The next line contains an integer, k, denoting the minimum sum of the subsequence elements Each line i of the N subsequence line (where 0<=i<N) contais an integer describing A[i] //approach 1 int functionName(int n,int k, vector<int>a) { int dp[n][k + 1]; for (int i = 0; i < n; i++) { for (int j = 0; j <= k; j++) { dp[i][j] = INT_MAX; } } dp[0][0] = 0; for (int i = 1; i <= n; i++) { for (int j = 0; j <= k; j++) { dp[i][j] = dp[i - 1][j]; if (j - a[i - 1] >= 0 && dp[i - 1][j - a[i - 1]] != INT_MAX && a[i - 1] >= a[i - 2]) { dp[i][j] = min(dp[i][j], dp[i - 1][j - a[i - 1]] + 1); } } } int res = INT_MAX; for (int i = 0; i <= k; i++) { if (dp[n][i] < INF) { res = min(res, dp[n][i]); } } if (res == INF) { return -1; } else { return res; } } Language c++ Infosys @codewithvirus @codewithvirus @codewithvirus https://bit.ly/infosys-SP-DSP

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You are given a string S with length N and a number K For every substring from s with length K we will get all permutations of this substring, This means that if k=3, all permutations of the substring aab are aab aba, baa. Count the total number of distinct strings in all permatetions of all substrings of S with length K, Since the answer may be very large return it modulo 1000000007 Notes: If we have the string ab, bb, ab, ba, there are 3 distinct strings ab, bb, and ba Input Format The first line contains an integer, N, denoting the lehorn of the given string. The next line contains an integer, K, denoting the length of each substring. The next line contains a string, S, denoting the given string int f(int n) { int r = 1; for (int i = 2; i <= n; i++) { r = (r * i) % MOD; } return r; } int functionName(int N,int K, string s) { int r = 0; for (int i = 0; i <= N - K; i++) { unordered_map c; for (int j = i; j < i + K; j++) { c[S[j]]++; } int p = 1; for (auto count : c) { p = (p * f(count.second)) % MOD; } r = (r + f(K) / p) % MOD; } return r; }

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+2

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Infosys C++ language @codewithvirus You are given a string S of length N. You can erase any substring of S of size <N. You need to return the lexicographically smallest remaining string after erasing a substring of S.

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// 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; } // @codewithvirus // @codewithvirus // @codewithvirus // @codewithvirus

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// Infosys //Given Array A having N integers and divisor K int morethanNbyK(vector<int> arr, int n, int k) { int x = n / k; int ans = 0; unordered_map<int, int> freq; for (auto i : arr) freq[i]++; for (auto i : freq) { if (i.second > x) { ans += i.first; } } return ans; } // @codewithvirus // @codewithvirus // @codewithvirus // @codewithvirus

10 559

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10 559

If you have Infosys SP and DSP exam then join our Infosys group 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 🔴🔴🔴🔴🔴🔴🔴🔴🔴🔴🔴 https://bit.ly/infosys-SP-DSP

10 559

If you have Infosys SP and DSP exam then join our Infosys group 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 🔴🔴🔴🔴🔴🔴🔴🔴🔴🔴🔴 https://bit.ly/infosys-SP-DSP

10 559

If you have Infosys SP and DSP exam then join our Infosys group 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 🔴🔴🔴🔴🔴🔴🔴🔴🔴🔴🔴 https://bit.ly/infosys-SP-DSP

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https://youtu.be/hU29GF5jpHM

10 559

If you have Infosys SP and DSP exam then join our Infosys group 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 🔴🔴🔴🔴🔴🔴🔴🔴🔴🔴🔴 https://bit.ly/infosys-SP-DSP

10 559

If you have Infosys SP and DSP exam then join our Infosys group 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 👉 @Infosyavirustech 👈 🔴🔴🔴🔴🔴🔴🔴🔴🔴🔴🔴 https://bit.ly/infosys-SP-DSP

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