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def count(a, X):     a.sort()     ans = 0     n = len(a)     for i in range(n):         if (i < n // 2):             ans += max(0, a[i] - X)         elif (i == n // 2):             ans += abs(X - a[i])         else:             ans += max(0, X - a[i]);     return ans IBM Telegram:-

int solve(vector& nums) {     vector s(nums);     sort(s.begin(), s.end());     int i = 0, j = nums.size() - 1;     while (i < nums.size() && nums[i] == s[i]) {         i++;     }     while (j > i && nums[j] == s[j]) {         j--;     }     return j - i + 1; } Arrange the heights Apple ✅ Telegram:-

void solve(string x){     ll n = x.size(); vector> dp(n,vector(n,0)); for(ll i=0;i

#include<bits/stdc++.h> using namespace std; void dfs(int node, vector<int>& vis, vector<vector<int>>& adj) {     vis[node] = 1;     for(auto it : adj[node]) {         if(!vis[it]) {             dfs(it, vis, adj);         }     } } int main() {     int n, m;     cin >> n >> m;     vector<vector<int>> adj(n);     for(int i = 0; i < n; i++) {         for(int j = 0; j < m; j++) {             int x;             cin >> x;             if(x == 1) {                 adj[i].push_back(j);                 adj[j].push_back(i);             }         }     }     vector<int> vis(n, 0);     int cc = 0;     for(int i = 0; i < n; i++) {         if(!vis[i]) {             dfs(i, vis, adj);             cc++;         }     }     cout << cc << endl;     return 0; }

int getMinOperations(vector arr) {           int n=arr.size();     int operations = 0;     for (int i = 1; i < n - 1; i++)     {         if (arr[i] < 0 && arr[i - 1] < 0)         {             arr[i] = 1e9;             operations++;         }         else if (arr[i] < 0 && arr[i - 1] > 0 && arr[i + 1] > 0)         {             if ((arr[i] + arr[i - 1] < 0) || (arr[i] + arr[i + 1] < 0))             {                 arr[i] = 1e9;                 operations++;             }         }     }    return operations;  } Make The Array Postive ✅

#include <bits/stdc++.h> using namespace std; int main() {     int n;     cin >> n;     vector<vector<int>> freq(n, vector<int> (26, 0));     string s;     for(int i = 0; i < n; i++) {         string s;         cin >> s;         for(auto &ch : s)               freq[i][ch - 'a']++;     }     long long ans = 0;     for(int i = 0; i < 26; i++) {         int minValue = INT_MAX;         for(int j = 0; j < n; j++)             minValue = min(minValue, freq[j][i]);         ans += minValue;     }     cout << ans << endl; } Service Now ✅

#include <bits/stdc++.h> #define ll long long using namespace std; ll solve(ll k, ll s) {     auto comb = [](ll n) {         return n * (n - 1) / 2;     };     if (k > 3 * s) return 0;     ll ans = comb(k + 2);     if (k > s) ans -= 3 * comb(k - s + 1);     if (k - 2 >= 2 * s) ans += 3 * comb(k - 2 * s);     return ans; } int main() {     ll n, s;     cin >> n >> s;     cout << solve(s,n) << endl;     return 0; } Distribute Car Toy Service Now ✅

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def decodeSequence(binarySequence): nucleobases = {'001': 'C', '011': 'A', '101': 'T','010': 'G', '110': 'U'} if len(binarySequence) % 3 != 0: print("Error: The length of the input string should be a multiple of 3.") return decodedSequence = "" for i in range(0, len(binarySequence), 3): chunk = binarySequence[i:i+3] if i == 0 and chunk == '000': identifier = 'DNA' elif i == 0 and chunk == '111': identifier = 'RNA' else: nucleobase = nucleobases.get(chunk, 'X') if nucleobase == 'U' and identifier == 'DNA': nucleobase = 'T' elif nucleobase == 'T' and identifier == 'RNA':                 nucleobase = 'U' decodedSequence += nucleobase print(decodedSequence) binaryInput = input("Enter the binary sequence: ") decodeSequence(binaryInput) DNA IBM Telegram:- @allcoding1_official

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string solve(string bs) {     map nb = {         {"001", "C"},         {"010", "G"},         {"011", "A"},         {"101", "T"},         {"110", "U"},         {"000", "DNA"},         {"111", "RNA"}     };     string ds = "";     string t = nb[bs.substr(0, 3)];     for(int i = 3; i < bs.length(); i += 3) {         string b = bs.substr(i, 3);         if(nb.find(b) != nb.end()) {             string x = nb[b];             if(t == "DNA" && x == "U") {                 x = "T";             }             ds += x;         } else {             ds += "Error";         }     }     return ds; } DNA✅ IBM

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bool isPal(int n) {     int r, s = 0, t;     t = n;     while (n > 0) {         r = n % 10;         s = (s * 10) + r;         n = n / 10;     }     return (t == s); } int firstPal(int n) {     int i = 1;     while (true) {         if (isPal(i)) {             int d = 1 + log10(i);             if (d == n)                 return i;         }         i++;     } } void login(int d, string u, string p) {     map users = {         {"user1", "pass1"},         {"user2", "pass2"},         {"user3", "pass3"},         {"user4", "pass4"},         {"user5", "pass5"}     };     if (users.find(u) != users.end() && users[u] == p) {         int t = firstPal(d);         cout << "Welcome " << u << " and the generated token is: token-" << t << endl;     } else {         cout << "UserId or password is not valid, please try again." << endl;     } } IBM✅

Here's a Python program to simulate the given problem: `python def print_terrain(terrain): for row in terrain: print(''.join(row)) def flow_water(terrain, n): water_level = int(terrain[n // 2][n // 2]) terrain[n // 2][n // 2] = 'W' def can_flow(x, y, direction): if direction == 'N': return x &gt; 0 and terrain[x-1][y] != 'W' and int(terrain[x-1][y]) &lt;= water_level elif direction == 'S': return x &lt; n - 1 and terrain[x+1][y] != 'W' and int(terrain[x+1][y]) &lt;= water_level elif direction == 'E': return y &lt; n - 1 and terrain[x][y+1] != 'W' and int(terrain[x][y+1]) &lt;= water_level elif direction == 'W': return y &gt; 0 and terrain[x][y-1] != 'W' and int(terrain[x][y-1]) &lt;= water_level def flow(x, y): if can_flow(x, y, 'N'): terrain[x-1][y] = 'W' return True if can_flow(x, y, 'S'): terrain[x+1][y] = 'W' return True if can_flow(x, y, 'E'): terrain[x][y+1] = 'W' return True if can_flow(x, y, 'W'): terrain[x][y-1] = 'W' return True return False while True: print_terrain(terrain) has_flown = False for i in range(n): for j in range(n): if terrain[i][j] == 'W': if flow(i, j): has_flown = True if not has_flown: water_level += 1 print(f"Cannot flow, increasing water level to {water_level}") break if any(cell == 'W' and (i == 0 or j == 0 or i == n - 1 or j == n - 1) for i, row in enumerate(terrain) for j, cell in enumerate(row)): print("Reached edge, exiting.") break n = 7 terrain = [ [494, 88, 89, 778, 984, 726, 587], [340, 959, 220, 301, 639, 280, 290], [666, 906, 632, 824, 127, 505, 787], [673, 499, 843, 172, 193, 613, 154], [544, 211, 124, 60, 575, 572, 389], [635, 170, 174, 946, 593, 314, 300], [620, 167, 931, 780, 416, 954, 275] ] flow_water(terrain, n)