I need help sorting a stack in DESCENDING ORDER so [44,32,21,11,0,-4,-5,-12,-24].
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I need help sorting a stack in DESCENDING ORDER so [44,32,21,11,0,-4,-5,-12,-24].
-----------------------------------------------------------------------------------------
Below is my failing code for [1,-1] where [-1,1] is being returned.
------------------------------------------------------------------------------------------
public Stack<Integer> solution(Stack<Integer> stack) {
if (!stack.isEmpty()) {
int top = stack.pop();
sort(stack);
sortUtil(stack, top);
}
return stack;
}
private static void sortUtil(Stack<Integer> stack, int data) {
if (stack.isEmpty() || stack.peek() < data) {
stack.push(data);
} else {
int top = stack.pop();
sortUtil(stack, data);
stack.push(top);
}
}
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- [JAVA ]I need help sorting a stack in DESCENDING ORDER so [44,32,21,11,0,-4,-5,-12,-24]. ----------------------------------------------------------------------------------------- Below is my failing code for [1,-1] where [-1,1] is being returned. ------------------------------------------------------------------------------------------ public Stack<Integer> solution(Stack<Integer> stack) {if (!stack.isEmpty()) {int top = stack.pop();sort(stack);sortUtil(stack, top);}return stack;}private static void sortUtil(Stack<Integer> stack, int data) {if (stack.isEmpty() || stack.peek() < data) {stack.push(data);} else {int top = stack.pop();sortUtil(stack, data);stack.push(top);}}Consider the following infix expression.( 5 + 8 ) * 9 – 7 * 10 + 9.Apply infix-to-postfix conversion algorithm to generate the correspondingpostfix expression from the given infix expression. You have to show thecontent of stack at each stage of conversion.Evaluate Infix + - * / 14 7 3 4 / 9 3 . Simulate it using Stack Algorithm. Please help sir. Thx.
- Observe the stack's reaction when given some space.Write a menu driven program as follows for STACK ---------------------Array based implementation of STACK----------------------------------- 1. Push an element on stack 2. Pop an element from stack 3. Display all 4. Top element 5. Exit ------------------------------------------------------------------------------------------------------------- Please Enter Your Choice: Note: use c++ language solve as soon as possibleDetermine how the stack performs when allowed to be itself.
- Java AddingLargeNumbers() Read the numerals of the first number and store the numbers corresponding to them on one stack Read the numerals of the second number and store the numbers to them on another stack; Carry = 0; While at least one stack is not empty Pop a number from each nonempty stack and add them to carry; Push the unit part on the result stack; Store the “carry over” in carry; One stack is empty, so a number is popped from the nonempty stack, added to the carry, and the result is stored on the resultStack. This must be repeated until the nonempty stack is empty. Push carry on the result stack if it is not zero; Pop numbers from the result stack and display them. Using javaDetermine the stack's performance when free.SOLVE THE MAZE FUNCTION USING STACK.PLEASE ONLY C CODE,DONT USE C++ AND C#. /* R\C 0 1 2 3 4 5 6 7 *//* 0 */ {0, 0, 0, 1, 0, 0, 0, 0},/* 1 */ {0, 1, 1, 1, 0, 0, 1, 0},/* 2 */ {0, 1, 0, 1, 0, 0, 1, 0},/* 3 */ {0, 1, 0, 1, 1, 1, 1, 0},/* 4 */ {0, 1, 0, 0, 0, 0, 1, 1},/* 5 */ {0, 1, 1, 0, 1, 1, 1, 0},/* 6 */ {0, 0, 0, 0, 1, 0, 0, 0},/* 7 */ {0, 0, 0, 0, 1, 0, 0, 0}, #include <stdio.h>#include <stdlib.h>#include <string.h> typedef struct LINKED_STACK_NODE_s *LINKED_STACK_NODE; typedef struct LINKED_STACK_NODE_s{LINKED_STACK_NODE next;void *data;} LINKED_STACK_NODE_t[1]; typedef struct LINKED_STACK_s{LINKED_STACK_NODE head;int count;} LINKED_STACK_t[1], *LINKED_STACK; typedef struct{int R;int C;} POS_t[1], *POS; LINKED_STACK stack_init();void stack_free(LINKED_STACK stack);void stack_push(LINKED_STACK stack, void *data);void *stack_pop(LINKED_STACK stack);void *stack_top(LINKED_STACK stack);int is_empty(LINKED_STACK stack); int is_empty(LINKED_STACK stack){return…
- Imbalanced arithmetic expressions are given asi) (A+B) * – (C+D+Fii) – ((A+B+C) * – (E+F)))The solution to the problem is an easy but elegant use of a stack to check formismatched parentheses. The general pseudo-code procedure for the problem is?Solve the maze function using stack.PLEASE ONLY C CODE, DONT USE C++ AND C# . /* R\C 0 1 2 3 4 5 6 7 *//* 0 */ {0, 0, 0, 1, 0, 0, 0, 0},/* 1 */ {0, 1, 1, 1, 0, 0, 1, 0},/* 2 */ {0, 1, 0, 1, 0, 0, 1, 0},/* 3 */ {0, 1, 0, 1, 1, 1, 1, 0},/* 4 */ {0, 1, 0, 0, 0, 0, 1, 1},/* 5 */ {0, 1, 1, 0, 1, 1, 1, 0},/* 6 */ {0, 0, 0, 0, 1, 0, 0, 0},/* 7 */ {0, 0, 0, 0, 1, 0, 0, 0}, void solve_maze(){int M[8][8] = {/* R\C 0 1 2 3 4 5 6 7 *//* 0 */ {0, 0, 0, 1, 0, 0, 0, 0},/* 1 */ {0, 1, 1, 1, 0, 0, 1, 0},/* 2 */ {0, 1, 0, 1, 0, 0, 1, 0},/* 3 */ {0, 1, 0, 1, 1, 1, 1, 0},/* 4 */ {0, 1, 0, 0, 0, 0, 1, 1},/* 5 */ {0, 1, 1, 0, 1, 1, 1, 0},/* 6 */ {0, 0, 0, 0, 1, 0, 0, 0},/* 7 */ {0, 0, 0, 0, 1, 0, 0, 0},};int I, J, R = 0, C = 3;LINKED_STACK stack;POS pos = (POS)malloc(sizeof(POS));int flag;stack = stack_init(); do{// TODO: Fill this block.} while (R != 7 && C != 7 && R != 0 && C != 0);}int main(){solve_maze();return 0;}Given a stack, switch_pairs function takes a stack as a parameter and thatswitches successive pairs of numbers starting at the bottom of the stack.For example, if the stack initially stores these values:bottom [3, 8, 17, 9, 1, 10] topYour function should switch the first pair (3, 8),the second pair (17, 9), ...:bottom [8, 3, 9, 17, 10, 1] topif there are an odd number of values in the stack, the value at the top of thestack is not moved: For example:bottom [3, 8, 17, 9, 1] topIt would again switch pairs of values, but the value at thetop of the stack (1)would not be movedbottom [8, 3, 9, 17, 1] topNote: There are 2 solutions:first_switch_pairs: it uses a single stack as auxiliary storagesecond_switch_pairs: it uses a single queue as auxiliary storage""".