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https://t.me/code_alphix/4871?single
- cognizant : https://t.me/code_alphix/4864
-capgemini:
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- Amazon
-Revature
-Wiley edge (mthree)
-Wipro : https://t.me/code_alphix/4859
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v, a, t = map(int, input().split())
u = v - a * t
print(u)
Initial velocity // wipro
import java.util.*;
public class Main {
static int countMatches(List<String> grid1, List<String> grid2) {
int count = 0;
int m = grid1.size();
int n = grid1.get(0).length();
int[][] g1 = new int[m][n];
int[][] g2 = new int[m][n];
int[][] orMap = new int[m][n];
for(int i=0; i<m; i++){
char[] str1 = grid1.get(i).toCharArray();
char[] str2 = grid2.get(i).toCharArray();
for(int j=0; j<n; j++){
g1[i][j] = str1[j] - '0';
g2[i][j] = str2[j] - '0';
}
}
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
orMap[i][j] = g1[i][j] | g2[i][j];
}
}
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
if(orMap[i][j] == 1 && dfs(g1, g2, orMap, i, j)) count++;
}
}
return count;
}
private static boolean dfs(int[][] g1, int[][] g2, int[][] orMap, int i, int j) {
int m = orMap.length;
int n = orMap[0].length;
orMap[i][j] = 0;
boolean up = true, down = true, left = true, right = true;
if(i > 0 && orMap[i-1][j] == 1) up = dfs(g1, g2, orMap, i-1, j);
if(i < m - 1 && orMap[i+1][j] == 1) down = dfs(g1, g2, orMap, i+1, j);
if(j > 0 && orMap[i][j-1] == 1) left = dfs(g1, g2, orMap, i, j-1);
if(j < n - 1 && orMap[i][j+1] == 1) right = dfs(g1, g2, orMap, i, j+1);
return g1[i][j] == 1 && g2[i][j] == 1 && up && down && left && right;
}
countmatches // LinkedIn 👍
import math
def closestColor(pixels):
pure_colors = {
"Black": (0, 0, 0),
"White": (255, 255, 255),
"Red": (255, 0, 0),
"Green": (0, 255, 0),
"Blue": (0, 0, 255)
}
def distance(c1, c2):
return sum((a - b) ** 2 for a, b in zip(c1, c2))
results = []
for pixel in pixels:
rgb = (int(pixel[i:i+8], 2) for i in (0, 8, 16))
closest = min(pure_colors.items(), key=lambda x: distance(rgb, x[1]))
min_dist = distance(rgb, closest[1])
ties = [name for name, color in pure_colors.items() if distance(rgb, color) == min_dist]
results.append("Ambiguous" if len(ties) > 1 else ties[0])
return results
Closestcolor// LinkedIn 👍
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