71. What will be the value stored in max_value when the following code is executed?
#include<stdio.h>
#include<limits.h>
int max_of_two(int a, int b)
{
if(a > b)
return a;
return b;
}
int rod_cut(int *prices, int len)
{
int max_price = INT_MIN; // INT_MIN is the min value an integer can take
int i;
if(len <= 0 )
return 0;
for(i = 0; i < len; i++)
// subtract 1 because index starts from 0
max_price = max_of_two(prices[i] + rod_cut(prices,len - i - 1), max_price);
return max_price;
}
int main()
{
int prices[]={2, 5, 6, 9, 9, 17, 17, 18, 20, 22},len_of_rod = 3;
int ans = rod_cut(prices, len_of_rod);
printf("%d",ans);
return 0;
}
#include<stdio.h>
#include<limits.h>
int max_of_two(int a, int b)
{
if(a > b)
return a;
return b;
}
int rod_cut(int *prices, int len)
{
int max_price = INT_MIN; // INT_MIN is the min value an integer can take
int i;
if(len <= 0 )
return 0;
for(i = 0; i < len; i++)
// subtract 1 because index starts from 0
max_price = max_of_two(prices[i] + rod_cut(prices,len - i - 1), max_price);
return max_price;
}
int main()
{
int prices[]={2, 5, 6, 9, 9, 17, 17, 18, 20, 22},len_of_rod = 3;
int ans = rod_cut(prices, len_of_rod);
printf("%d",ans);
return 0;
}72. What will be the value stored in arr[2][2] when the following code is executed?
#include<stdio.h>
int get_ways(int num_of_dice, int num_of_faces, int S)
{
int arr[num_of_dice + 1][S + 1];
int dice, face, sm;
for(dice = 0; dice <= num_of_dice; dice++)
for(sm = 0; sm <= S; sm++)
arr[dice][sm] = 0;
for(sm = 1; sm <= S; sm++)
arr[1][sm] = 1;
for(dice = 2; dice <= num_of_dice; dice++)
{
for(sm = 1; sm <= S; sm++)
{
for(face = 1; face <= num_of_faces && face < sm; face++)
arr[dice][sm] += arr[dice - 1][sm - face];
}
}
return arr[num_of_dice][S];
}
int main()
{
int num_of_dice = 3, num_of_faces = 4, sum = 6;
int ans = get_ways(num_of_dice, num_of_faces, sum);
printf("%d",ans);
return 0;
}
#include<stdio.h>
int get_ways(int num_of_dice, int num_of_faces, int S)
{
int arr[num_of_dice + 1][S + 1];
int dice, face, sm;
for(dice = 0; dice <= num_of_dice; dice++)
for(sm = 0; sm <= S; sm++)
arr[dice][sm] = 0;
for(sm = 1; sm <= S; sm++)
arr[1][sm] = 1;
for(dice = 2; dice <= num_of_dice; dice++)
{
for(sm = 1; sm <= S; sm++)
{
for(face = 1; face <= num_of_faces && face < sm; face++)
arr[dice][sm] += arr[dice - 1][sm - face];
}
}
return arr[num_of_dice][S];
}
int main()
{
int num_of_dice = 3, num_of_faces = 4, sum = 6;
int ans = get_ways(num_of_dice, num_of_faces, sum);
printf("%d",ans);
return 0;
}73. Which of the following problems is NOT solved using dynamic programming?
74. What is the time complexity of the following dynamic programming algorithm used to find the maximum sub-array sum?
#include<stdio.h>
int max_num(int a,int b)
{
if(a> b)
return a;
return b;
}
int maximum_subarray_sum(int *arr, int len)
{
int sum[len], idx;
sum[0] = arr[0];
for(idx = 1; idx < len; idx++)
sum[idx] = max_num(sum[idx - 1] + arr[idx], arr[idx]);
int mx = sum[0];
for(idx = 0; idx < len; idx++)
if(sum[idx] > mx)
mx =sum[idx];
return mx;
}
int main()
{
int arr[] = {-2, -5, 6, -2, 3, -1, 0,-5, 6}, len = 9;
int ans = maximum_subarray_sum(arr, len);
printf("%d",ans);
return 0;
}
#include<stdio.h>
int max_num(int a,int b)
{
if(a> b)
return a;
return b;
}
int maximum_subarray_sum(int *arr, int len)
{
int sum[len], idx;
sum[0] = arr[0];
for(idx = 1; idx < len; idx++)
sum[idx] = max_num(sum[idx - 1] + arr[idx], arr[idx]);
int mx = sum[0];
for(idx = 0; idx < len; idx++)
if(sum[idx] > mx)
mx =sum[idx];
return mx;
}
int main()
{
int arr[] = {-2, -5, 6, -2, 3, -1, 0,-5, 6}, len = 9;
int ans = maximum_subarray_sum(arr, len);
printf("%d",ans);
return 0;
}75. What is the output of the following code?
#include<stdio.h>
#include<string.h>
int max(int a, int b)
{
if(a > b)
return a;
return b;
}
int min_ins(char *s)
{
int len = strlen(s), i, j;
int arr[len + 1][len + 1];
char rev[len + 1];
strcpy(rev, s);
strrev(rev);
for(i = 0;i <= len; i++)
arr[i][0] = 0;
for(i = 0; i <= len; i++)
arr[0][i] = 0;
for(i = 1; i <= len; i++)
{
for(j = 1; j <= len; j++)
{
if(s[i - 1] == rev[j - 1])
arr[i][j] = arr[i - 1][j - 1] + 1;
else
arr[i][j] = max(arr[i - 1][j], arr[i][j - 1]);
}
}
return len - arr[len][len];
}
int main()
{
char s[] = "efgfe";
int ans = min_ins(s);
printf("%d",ans);
return 0;
}
#include<stdio.h>
#include<string.h>
int max(int a, int b)
{
if(a > b)
return a;
return b;
}
int min_ins(char *s)
{
int len = strlen(s), i, j;
int arr[len + 1][len + 1];
char rev[len + 1];
strcpy(rev, s);
strrev(rev);
for(i = 0;i <= len; i++)
arr[i][0] = 0;
for(i = 0; i <= len; i++)
arr[0][i] = 0;
for(i = 1; i <= len; i++)
{
for(j = 1; j <= len; j++)
{
if(s[i - 1] == rev[j - 1])
arr[i][j] = arr[i - 1][j - 1] + 1;
else
arr[i][j] = max(arr[i - 1][j], arr[i][j - 1]);
}
}
return len - arr[len][len];
}
int main()
{
char s[] = "efgfe";
int ans = min_ins(s);
printf("%d",ans);
return 0;
}76. What is the space complexity of the following dynamic programming implementation used to find the minimum number of jumps?
#include<stdio.h>
#include<limits.h>
int min_jump(int *arr, int len)
{
int j, idx, jumps[len];
jumps[len - 1] = 0;
for(idx = len - 2; idx >= 0; idx--)
{
int tmp_min = INT_MAX;
for(j = 1; j <= arr[idx] && idx + j < len; j++)
{
if(jumps[idx + j] + 1 < tmp_min)
tmp_min = jumps[idx + j] + 1;
}
jumps[idx] = tmp_min;
}
return jumps[0];
}
int main()
{
int arr[] ={1, 1, 1, 1, 1, 1, 1, 1, 1},len = 9;
int ans = min_jump(arr,len);
printf("%d\n",ans);
return 0;
}
#include<stdio.h>
#include<limits.h>
int min_jump(int *arr, int len)
{
int j, idx, jumps[len];
jumps[len - 1] = 0;
for(idx = len - 2; idx >= 0; idx--)
{
int tmp_min = INT_MAX;
for(j = 1; j <= arr[idx] && idx + j < len; j++)
{
if(jumps[idx + j] + 1 < tmp_min)
tmp_min = jumps[idx + j] + 1;
}
jumps[idx] = tmp_min;
}
return jumps[0];
}
int main()
{
int arr[] ={1, 1, 1, 1, 1, 1, 1, 1, 1},len = 9;
int ans = min_jump(arr,len);
printf("%d\n",ans);
return 0;
}77. What is the output of the following code?
#include<stdio.h>
#include<string.h>
int get_min(int a, int b)
{
if(a < b)
return a;
return b;
}
int edit_distance(char *s1, char *s2)
{
int len1,len2,i,j,min;
len1 = strlen(s1);
len2 = strlen(s2);
int arr[len1 + 1][len2 + 1];
for(i = 0;i <= len1; i++)
arr[i][0] = i;
for(i = 0; i <= len2; i++)
arr[0][i] = i;
for(i = 1; i <= len1; i++)
{
for(j = 1; j <= len2; j++)
{
min = get_min(arr[i-1][j],arr[i][j-1]) + 1;
if(s1[i - 1] == s2[j - 1])
{
if(arr[i-1][j-1] < min)
min = arr[i-1][j-1];
}
else
{
if(arr[i-1][j-1] + 1 < min)
min = arr[i-1][j-1] + 1;
}
arr[i][j] = min;
}
}
return arr[len1][len2];
}
int main()
{
char s1[] = "abcd", s2[] = "dcba";
int ans = edit_distance(s1, s2);
printf("%d",ans);
return 0;
}
#include<stdio.h>
#include<string.h>
int get_min(int a, int b)
{
if(a < b)
return a;
return b;
}
int edit_distance(char *s1, char *s2)
{
int len1,len2,i,j,min;
len1 = strlen(s1);
len2 = strlen(s2);
int arr[len1 + 1][len2 + 1];
for(i = 0;i <= len1; i++)
arr[i][0] = i;
for(i = 0; i <= len2; i++)
arr[0][i] = i;
for(i = 1; i <= len1; i++)
{
for(j = 1; j <= len2; j++)
{
min = get_min(arr[i-1][j],arr[i][j-1]) + 1;
if(s1[i - 1] == s2[j - 1])
{
if(arr[i-1][j-1] < min)
min = arr[i-1][j-1];
}
else
{
if(arr[i-1][j-1] + 1 < min)
min = arr[i-1][j-1] + 1;
}
arr[i][j] = min;
}
}
return arr[len1][len2];
}
int main()
{
char s1[] = "abcd", s2[] = "dcba";
int ans = edit_distance(s1, s2);
printf("%d",ans);
return 0;
}78. Wagner-Fischer algorithm is used to find . . . . . . . .
79. What is the length of the longest palindromic subsequence for the string "ababcdabba"?
80. The Knapsack problem is an example of . . . . . . . .
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