Infosys Coding Solution 29 July 3 PM Slot

Infosys Coding Solution 29 July 3 PM Slot

Infosys Code 1

Number Of Subsequences Code For Infosys

int A[] = 10,13,7,8,14,11;
int n = 6;

int memo[6];//initialized with -1s;

int count(int currIndex)
if (currIndex == n-1) return 1;
if (memo[currIndex] != -1) return memo[currIndex];

int res = 0;
for (int i=currIndex+1 ; i<n ; i++)
if (abss(A[currIndex] – A[i]) <= 3)
res += count(i);

memo[currIndex] = res;
return res;

c++

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Infosys Code 2

def minOperations(arr1, arr2, i, j):

# Base Case

if arr1 == arr2:

return 0

if i >= len(arr1) or j >= len(arr2):

return 0

# If arr[i] < arr[j]

if arr1[i] < arr2[j]:

# Include the current element

return 1 \

+ minOperations(arr1, arr2, i + 1, j + 1)

# Otherwise, excluding the current element

return max(minOperations(arr1, arr2, i, j + 1),

minOperations(arr1, arr2, i + 1, j))

# Function that counts the minimum

# moves required to sort the array

def minOperationsUtil(arr):

brr = sorted(arr);

# If both the arrays are equal

if(arr == brr):

# No moves required

print(“0”)

# Otherwise

else:

# Print minimum operations required

print(minOperations(arr, brr,)

Python

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Infosys Code 3

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C++ language
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Infosys Code 4

# to make a given change V

import sys

# m is size of coins array (number of different coins)

def minCoins(coins, m, V):

# base case

if (V == 0):

return 0

# Initialize result

res = sys.maxsize

# Try every coin that has smaller value than V

for i in range(0, m):

if (coins[i] <= V):

sub_res = minCoins(coins, m, V-coins[i])

# Check for INT_MAX to avoid overflow and see if

# result can minimized

if (sub_res != sys.maxsize and sub_res + 1 < res):

res = sub_res + 1

return res

# Driver program to test above function

coins = [9, 6, 5, 1]

m = len(coins)

V = 11

print(“Minimum coins required is”,minCoins(coins, m, V))

Python 3
program to find minimum of coins Code

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Infosys Code 5

C++
Lexographically smallest question codeINFOSYS EXAM CODE

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Infosys Code 6

array AR of size N

perfect_sum(arr, s, result) :
x = [0]*len(arr)
j = len(arr) – 1

while (s > 0) :
x[j] = s % 2
s = s // 2
j -= 1
sum = 0
for i in range(len(arr)) :
if (x[i] == 1) :
sum += arr[i]
if (sum == result) :
print(““,end=””);
for i in range(len(arr)) :
if (x[i] == 1) :
print(arr[i],end= “, “);
print(“, “,end=””)

def print_subset(arr, K) :
x = pow(2, len(arr))
for i in range(1, x):
perfect_sum(arr, i, K)

# Driver code
arr = [ ]
n = int(input(“Enter length of array : “))
s=int(input(“Enter sum : “))
for i in range(n):
ele=int(input(“Enter element : “))
arr.append(ele)
print_subset(arr, s)

Python

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Infosys Code 7