Cognizant Hiring Software Engineer:
Graduation Year: 2023 / 2024 / 2025 / 2026
Salary: Upto 7.5 LPA
Location: Bangalore
Apply Now:- https://www.jobseekers19.com/2026/07/congnizan-hiring-for-software-engineer.html
Graduation Year: 2023 / 2024 / 2025 / 2026
Salary: Upto 7.5 LPA
Location: Bangalore
Apply Now:- https://www.jobseekers19.com/2026/07/congnizan-hiring-for-software-engineer.html
❤1
KPMG India Hiring
Role: Executive – Python
Job ID: INTG10045780
Posted On: 13 July 2026
Required Skill: Python Development
Apply Now:-https://www.jobseekers19.com/2026/07/kpmg-india-hiring-2026.html
Role: Executive – Python
Job ID: INTG10045780
Posted On: 13 July 2026
Required Skill: Python Development
Apply Now:-https://www.jobseekers19.com/2026/07/kpmg-india-hiring-2026.html
Deloitte is Hiring for Services Support Associate
Apply Now:- https://www.jobseekers19.com/2026/07/deloitte-is-hiring-for-services-support.html
Apply Now:- https://www.jobseekers19.com/2026/07/deloitte-is-hiring-for-services-support.html
Jobseekers19
Deloitte is Hiring for Services Support Associate I
Services Support Associate I - Talent Processes and Business Support - Talent Services - LDS ( Instructional Design) - Hyderabad ...
Google is hiring
Position: Data Analytics Apprenticeship
Qualifications: Bachelor’s degree
Experience: 0-1 (Years)
Location: Hyderabad; Bengaluru; Gurgaon; Mumbai, India
Apply Now: https://www.google.com/about/careers/applications/jobs/results/114421121054319302-data-analytics-apprenticeship-march-2027-start
Position: Data Analytics Apprenticeship
Qualifications: Bachelor’s degree
Experience: 0-1 (Years)
Location: Hyderabad; Bengaluru; Gurgaon; Mumbai, India
Apply Now: https://www.google.com/about/careers/applications/jobs/results/114421121054319302-data-analytics-apprenticeship-march-2027-start
❤2
𝗖𝗼𝗴𝗻𝗶𝘇𝗮𝗻𝘁 𝗙𝗿𝗲𝘀𝗵𝗲𝗿 𝗛𝗶𝗿𝗶𝗻𝗴 – Digital Workplace Practice (Service Desk)
𝗘𝗹𝗶𝗴𝗶𝗯𝗶𝘁𝘆: 2025 & 2026 Graduates (3-Year UG Degree – B.Sc, BCA, B.Com, BA, BBA, B.Voc, BMS, etc.)
𝗥𝗼𝗹𝗲: Service Desk (Digital Workplace Practice)
𝗟𝗼𝗰𝗮𝘁𝗶𝗼𝗻: PAN India
𝗦𝗮𝗹𝗮𝗿𝘆: As Per Company Standards
𝗟𝗮𝘀𝘁 𝗗𝗮𝘁𝗲: 31st July 2026
𝗢𝗙𝗙𝗜𝗖𝗜𝗔𝗟 𝗔𝗣𝗣𝗟𝗜𝗖𝗔𝗧𝗜𝗢𝗡 𝗟𝗜𝗡𝗞: https://app.joinsuperset.com/join/#/signup/student/jobprofiles/a8af19f3-444e-43bd-8828-e8656f637a17
𝗘𝗹𝗶𝗴𝗶𝗯𝗶𝘁𝘆: 2025 & 2026 Graduates (3-Year UG Degree – B.Sc, BCA, B.Com, BA, BBA, B.Voc, BMS, etc.)
𝗥𝗼𝗹𝗲: Service Desk (Digital Workplace Practice)
𝗟𝗼𝗰𝗮𝘁𝗶𝗼𝗻: PAN India
𝗦𝗮𝗹𝗮𝗿𝘆: As Per Company Standards
𝗟𝗮𝘀𝘁 𝗗𝗮𝘁𝗲: 31st July 2026
𝗢𝗙𝗙𝗜𝗖𝗜𝗔𝗟 𝗔𝗣𝗣𝗟𝗜𝗖𝗔𝗧𝗜𝗢𝗡 𝗟𝗜𝗡𝗞: https://app.joinsuperset.com/join/#/signup/student/jobprofiles/a8af19f3-444e-43bd-8828-e8656f637a17
❤1
Company:- Concentrix
Apply link :- https://concentrix.myamcat.com/register?data=zfmPuUF9UxJRR0maCTAXsacw8z2SCN9CHKUEKlBxfKKuOiCtWiv9x2nQTjNAj9LSeNqPICO7o0HQzAYmoS6GUA%3D%3D&utm_source=sp_auto_dm&utm_referrer=sp_auto_dm&fbclid=PAT01DUAP8tFxleHRuA2FlbQIxMABzcnRjBmFwcF9pZA81NjcwNjczNDMzNT
Test mail confirm ( must apply )
Follow For more jobs vecancy
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HCLTech Freshers Hiring
Eligible Batches: 2020–2026
Skills: Java, Python, C/C++, SQL, HTML, CSS, JavaScript, Cloud, Networking, Testing and Communication Skillls
Direct Test
Select a role based on the job description
Apply as soon as possible
Apply Here:
https://freshers.hcltech.com/
Eligible Batches: 2020–2026
Skills: Java, Python, C/C++, SQL, HTML, CSS, JavaScript, Cloud, Networking, Testing and Communication Skillls
Direct Test
Select a role based on the job description
Apply as soon as possible
Apply Here:
https://freshers.hcltech.com/
#include <bits/stdc++.h>
using namespace std;
long long solve(int N, int S, vector<int>& M) {
unordered_map<long long, long long> freq;
for (int x : M) :
long long ans = 0;
for (auto &[x, cnt] : freq) {
long long y = (long long)S - 2LL * x;
if (y < x) continue;
auto it = freq.find(y);
if (it == freq.end()) continue;
if (x == y)
ans += cnt * (cnt - 1) / 2;
else
ans += cnt * it->second;
}
return ans;
}
Barycentric Tether calibration ✅
C++
using namespace std;
long long solve(int N, int S, vector<int>& M) {
unordered_map<long long, long long> freq;
for (int x : M) :
long long ans = 0;
for (auto &[x, cnt] : freq) {
long long y = (long long)S - 2LL * x;
if (y < x) continue;
auto it = freq.find(y);
if (it == freq.end()) continue;
if (x == y)
ans += cnt * (cnt - 1) / 2;
else
ans += cnt * it->second;
}
return ans;
}
Barycentric Tether calibration ✅
C++
from collections import defaultdict
n = int(raw_input())
L = int(raw_input())
K = int(raw_input())
pts = list(map(int, raw_input().split()))
groups = defaultdict(list)
for x in pts:
groups[x % K].append(x)
ans = 0
for arr in groups.values():
arr.sort()
left = 0
for right in range(len(arr)):
while arr[right] - arr[left] > L:
left += 1
ans = max(ans, right - left + 1)
print code
All test cases pass
Python
Telegram:- @allcoding1_official
n = int(raw_input())
L = int(raw_input())
K = int(raw_input())
pts = list(map(int, raw_input().split()))
groups = defaultdict(list)
for x in pts:
groups[x % K].append(x)
ans = 0
for arr in groups.values():
arr.sort()
left = 0
for right in range(len(arr)):
while arr[right] - arr[left] > L:
left += 1
ans = max(ans, right - left + 1)
print code
All test cases pass
Python
Telegram:- @allcoding1_official
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
int solve(int N) {
vector<array<long long, 3>> dp(N + 1);
// last = 0 -> 1
// last = 1 -> 3
// last = 2 -> 4
for (int s = 1; s <= N; s++) {
if (s == 1) {
dp[s][0] = 1;
} else {
long long ways = 0;
if (s >= 1) {
ways = (dp[s - 1][1] + dp[s - 1][2]) % MOD;
}
dp[s][0] = ways;
}
if (s == 3) {
dp[s][1] = (dp[s][1] + 1) % MOD;
}
if (s > 3) {
dp[s][1] = (dp[s - 3][0] + dp[s - 3][2]) % MOD;
}
if (s == 4) {
dp[s][2] = (dp[s][2] + 1) % MOD;
}
if (s > 4) {
dp[s][2] = (dp[s - 4][0] + dp[s - 4][1]) % MOD;
}
}
return (dp[N][0] + dp[N][1] + dp[N][2]) % MOD;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int N;
cin >> N;
cout << solve(N) << '\n';
return 0;
}
using namespace std;
const int MOD = 1000000007;
int solve(int N) {
vector<array<long long, 3>> dp(N + 1);
// last = 0 -> 1
// last = 1 -> 3
// last = 2 -> 4
for (int s = 1; s <= N; s++) {
if (s == 1) {
dp[s][0] = 1;
} else {
long long ways = 0;
if (s >= 1) {
ways = (dp[s - 1][1] + dp[s - 1][2]) % MOD;
}
dp[s][0] = ways;
}
if (s == 3) {
dp[s][1] = (dp[s][1] + 1) % MOD;
}
if (s > 3) {
dp[s][1] = (dp[s - 3][0] + dp[s - 3][2]) % MOD;
}
if (s == 4) {
dp[s][2] = (dp[s][2] + 1) % MOD;
}
if (s > 4) {
dp[s][2] = (dp[s - 4][0] + dp[s - 4][1]) % MOD;
}
}
return (dp[N][0] + dp[N][1] + dp[N][2]) % MOD;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int N;
cin >> N;
cout << solve(N) << '\n';
return 0;
}
#include <bits/stdc++.h>
using namespace std;
int solve(int N, int K, vector<vector<int>>& cost, vector<int>& fatigue) {
const long long INF = 4e18;
vector<vector<long long>> prev(K, vector<long long>(2, INF));
vector<vector<long long>> cur(K, vector<long long>(2, INF));
for (int c = 0; c < K; c++)
prev[c][1] = cost[0][c];
for (int i = 1; i < N; i++) {
for (int c = 0; c < K; c++) {
cur[c][0] = cur[c][1] = INF;
}
for (int last = 0; last < K; last++) {
for (int streak = 1; streak <= 2; streak++) {
if (prev[last][streak - 1] == INF) continue;
for (int now = 0; now < K; now++) {
if (now == last) {
if (streak == 2) continue;
cur[now][1] = min(
cur[now][1],
prev[last][streak - 1] +
cost[i][now] + fatigue[now]
);
using namespace std;
int solve(int N, int K, vector<vector<int>>& cost, vector<int>& fatigue) {
const long long INF = 4e18;
vector<vector<long long>> prev(K, vector<long long>(2, INF));
vector<vector<long long>> cur(K, vector<long long>(2, INF));
for (int c = 0; c < K; c++)
prev[c][1] = cost[0][c];
for (int i = 1; i < N; i++) {
for (int c = 0; c < K; c++) {
cur[c][0] = cur[c][1] = INF;
}
for (int last = 0; last < K; last++) {
for (int streak = 1; streak <= 2; streak++) {
if (prev[last][streak - 1] == INF) continue;
for (int now = 0; now < K; now++) {
if (now == last) {
if (streak == 2) continue;
cur[now][1] = min(
cur[now][1],
prev[last][streak - 1] +
cost[i][now] + fatigue[now]
);
❤1
#include <bits/stdc++.h>
using namespace std;
long long solve(int N, vector<int>& target) {
long long total = 0;
for (int i = 0; i < N; i++) total += target[i];
long long sumX = 0;
long long carry = 0;
for (int i = 0; i + 1 < N; i++) {
long long cap = min((long long)target[i] - carry, (long long)target[i + 1]);
if (cap < 0) cap = 0;
sumX += cap;
carry = cap;
}
return total - sumX;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int N; cin >> N;
vector<int> target(N);
for (int i = 0; i < N; i++) cin >> target[i];
auto result = solve(N, target);
cout << result << endl;
return 0;
}
using namespace std;
long long solve(int N, vector<int>& target) {
long long total = 0;
for (int i = 0; i < N; i++) total += target[i];
long long sumX = 0;
long long carry = 0;
for (int i = 0; i + 1 < N; i++) {
long long cap = min((long long)target[i] - carry, (long long)target[i + 1]);
if (cap < 0) cap = 0;
sumX += cap;
carry = cap;
}
return total - sumX;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int N; cin >> N;
vector<int> target(N);
for (int i = 0; i < N; i++) cin >> target[i];
auto result = solve(N, target);
cout << result << endl;
return 0;
}
def solve(N: int, t: list) -> list:
if all(t[k] <= t[k+1] for k in range(N - 1)):
return [0, sum(t)]
p = next(k for k in range(N - 1) if t[k] > t[k+1])
M = [0] * N
M[N-1] = t[N-1]
for i in range(N - 2, -1, -1):
M[i] = min(t[i], M[i+1])
prefix_sum = [0] * (N + 1)
for i in range(N):
prefix_sum[i+1] = prefix_sum[i] + t[i]
best = -1
for i in range(0, p + 2):
if i > 0 and t[i-1] > M[i]:
continue
total = prefix_sum[i] + M[i] * (N - i)
best = max(best, total)
return [1, best]
if all(t[k] <= t[k+1] for k in range(N - 1)):
return [0, sum(t)]
p = next(k for k in range(N - 1) if t[k] > t[k+1])
M = [0] * N
M[N-1] = t[N-1]
for i in range(N - 2, -1, -1):
M[i] = min(t[i], M[i+1])
prefix_sum = [0] * (N + 1)
for i in range(N):
prefix_sum[i+1] = prefix_sum[i] + t[i]
best = -1
for i in range(0, p + 2):
if i > 0 and t[i-1] > M[i]:
continue
total = prefix_sum[i] + M[i] * (N - i)
best = max(best, total)
return [1, best]
#include <bits/stdc++.h>
using namespace std;
long long solve(int N, int M, long long K, long long C, vector<vector<int>>& g) {
vector<vector<long long>> dist(N, vector<long long>(M, LLONG_MAX));
dist[0][0] = g[0][0];
priority_queue<tuple<long long,int,int>, vector<tuple<long long,int,int>>, greater<>> pq;
pq.push({dist[0][0], 0, 0});
while (!pq.empty()) {
auto [d, r, c] = pq.top(); pq.pop();
if (d > dist[r][c]) continue;
if (r == N-1 && c == M-1) break;
for (int j = c+1; j < M; j++) {
long long diff = llabs((long long)g[r][j] - g[r][c]);
long long step = j - c;
long long cost;
if (step == 1) {
cost = g[r][j] + (diff > K ? C : 0);
} else {
if (diff > K) continue;
cost = g[r][j] + C * (step - 1);
}
long long nd = d + cost;
if (nd < dist[r][j]) { dist[r][j] = nd; pq.push({nd, r, j}); }
}
for (int i = r+1; i < N; i++) {
long long diff = llabs((long long)g[i][c] - g[r][c]);
long long step = i - r;
long long cost;
if (step == 1) {
cost = g[i][c] + (diff > K ? C : 0);
} else {
if (diff > K) continue;
cost = g[i][c] + C * (step - 1);
}
long long nd = d + cost;
if (nd < dist[i][c]) { dist[i][c] = nd; pq.push({nd, i, c}); }
}
}
return dist[N-1][M-1];
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int N; cin >> N;
int M; cin >> M;
long long K; cin >> K;
long long C; cin >> C;
vector<vector<int>> g(N, vector<int>(M));
for (int i = 0; i < N; i++)
for (int j = 0; j < M; j++) cin >> g[i][j];
auto result = solve(N, M, K, C, g);
cout << result << endl;
return 0;
}
using namespace std;
long long solve(int N, int M, long long K, long long C, vector<vector<int>>& g) {
vector<vector<long long>> dist(N, vector<long long>(M, LLONG_MAX));
dist[0][0] = g[0][0];
priority_queue<tuple<long long,int,int>, vector<tuple<long long,int,int>>, greater<>> pq;
pq.push({dist[0][0], 0, 0});
while (!pq.empty()) {
auto [d, r, c] = pq.top(); pq.pop();
if (d > dist[r][c]) continue;
if (r == N-1 && c == M-1) break;
for (int j = c+1; j < M; j++) {
long long diff = llabs((long long)g[r][j] - g[r][c]);
long long step = j - c;
long long cost;
if (step == 1) {
cost = g[r][j] + (diff > K ? C : 0);
} else {
if (diff > K) continue;
cost = g[r][j] + C * (step - 1);
}
long long nd = d + cost;
if (nd < dist[r][j]) { dist[r][j] = nd; pq.push({nd, r, j}); }
}
for (int i = r+1; i < N; i++) {
long long diff = llabs((long long)g[i][c] - g[r][c]);
long long step = i - r;
long long cost;
if (step == 1) {
cost = g[i][c] + (diff > K ? C : 0);
} else {
if (diff > K) continue;
cost = g[i][c] + C * (step - 1);
}
long long nd = d + cost;
if (nd < dist[i][c]) { dist[i][c] = nd; pq.push({nd, i, c}); }
}
}
return dist[N-1][M-1];
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int N; cin >> N;
int M; cin >> M;
long long K; cin >> K;
long long C; cin >> C;
vector<vector<int>> g(N, vector<int>(M));
for (int i = 0; i < N; i++)
for (int j = 0; j < M; j++) cin >> g[i][j];
auto result = solve(N, M, K, C, g);
cout << result << endl;
return 0;
}
❤1
import sys
input = sys.stdin.readline
def digit_sum(n: int) -> int:
return sum(int(c) for c in str(n))
def solve(N: int, S: int) -> list:
if S == 0:
m = 0
else:
num_digits = (S + 8) // 9 # ceil(S/9)
first = S - 9 * (num_digits - 1)
m = int(str(first) + '9' * (num_digits - 1))
if m <= N:
return [1, m]
else:
return [0, -1]
if name == "main":
try:
N = int(input())
S = int(input())
result = solve(N, S)
print(" ".join(map(str, result)))
except (EOFError, ValueError):
pass
input = sys.stdin.readline
def digit_sum(n: int) -> int:
return sum(int(c) for c in str(n))
def solve(N: int, S: int) -> list:
if S == 0:
m = 0
else:
num_digits = (S + 8) // 9 # ceil(S/9)
first = S - 9 * (num_digits - 1)
m = int(str(first) + '9' * (num_digits - 1))
if m <= N:
return [1, m]
else:
return [0, -1]
if name == "main":
try:
N = int(input())
S = int(input())
result = solve(N, S)
print(" ".join(map(str, result)))
except (EOFError, ValueError):
pass