Top 50 Most Asked C Programming with DSA Interview Questions

Read character by character with fgetc. FILE *fp = fopen(path, "r"); if (!fp) { perror("fopen"); return 1; } long lines = 0, words = 0, chars = 0; int c, inWord = 0; while ((c = fgetc(fp)) != EOF) { chars++; if (c == '\n') lines++; if (isspace(c)) inWord = 0; else if (!inWord) { inWord = 1; words++; } } fclose(fp); printf("%ld %ld %ld\n", lines, words, chars); c is int so EOF can be distinguished from valid bytes.

A pointer to pointer handles the empty list without special cases. int append(Node **head, int value) { Node *n = malloc(sizeof *n); if (!n) return -1; n->data = value; n->next = NULL; while (*head) head = &(*head)->next; *head = n; return 0; } Call it as append(&head, 5). Keep a tail pointer if you append often.

Use bit masks with shift operators. x |= (1u << n); // set bit n x &= ~(1u << n); // clear bit n x ^= (1u << n); // toggle bit n int bit = (x >> n) & 1u; // read bit n Use unsigned types so shifts are well defined. These operations are common in embedded code for configuring hardware registers.

Track the best sum ending here and the overall best. long maxSubarray(const int *a, int n) { long cur = a[0], best = a[0]; for (int i = 1; i < n; i++) { cur = (cur + a[i] > a[i]) ? cur + a[i] : a[i]; if (cur > best) best = cur; } return best; } Input {-2, 1, -3, 4, -1, 2, 1, -5, 4} returns 6 for subarray 4, -1, 2, 1.

Push digits and apply operators to the top two values. int evalPostfix(const char *s) { int st[100], top = -1; for (; *s; s++) { if (isdigit((unsigned char)*s)) st[++top] = *s - '0'; else if (strchr("+-*/", *s)) { int b = st[top--], a = st[top--]; switch (*s) { case '+': st[++top] = a + b; break; case '-': st[++top] = a - b; break; case '*': st[++top] = a * b; break; case '/': st[++top] = a / b; break; } } } return st[top]; } 23*5+ evaluates to 11.

Brian Kernighan's trick clears the lowest set bit each step. int countBits(unsigned int n) { int count = 0; while (n) { n &= n - 1; count++; } return count; } It runs once per set bit. GCC and Clang also offer __builtin_popcount(n), which uses a CPU instruction when available.

Use three reversals. void rev(int *a, int l, int r) { while (l < r) { int t = a[l]; a[l++] = a[r]; a[r--] = t; } } void rotate(int *a, int n, int k) { k %= n; if (k == 0) return; rev(a, 0, n - 1); rev(a, 0, k - 1); rev(a, k, n - 1); } For 1 2 3 4 5 with k = 2 the result is 4 5 1 2 3.

Sliding window with last seen positions. int longestUnique(const char *s) { int last[256]; for (int i = 0; i < 256; i++) last[i] = -1; int best = 0, start = 0; for (int i = 0; s[i]; i++) { unsigned char c = s[i]; if (last[c] >= start) start = last[c] + 1; last[c] = i; if (i - start + 1 > best) best = i - start + 1; } return best; } For abcabcbb it returns 3.

Flip each next pointer while walking the list. typedef struct Node { int data; struct Node *next; } Node; Node *reverse(Node *head) { Node *prev = NULL, *cur = head; while (cur) { Node *next = cur->next; cur->next = prev; prev = cur; cur = next; } return prev; } Time O(n), space O(1). The returned prev is the new head.

Sum of digits raised to the number of digits equals the number. int isArmstrong(int n) { int digits = 0, t = n, sum = 0; while (t) { digits++; t /= 10; } t = n; while (t) { int d = t % 10, p = 1; for (int i = 0; i < digits; i++) p *= d; sum += p; t /= 10; } return sum == n; } For example 153 = 1 + 125 + 27.

A header guard stops a header from being included twice in one translation unit, which would cause redefinition errors. #ifndef MATH_UTILS_H #define MATH_UTILS_H int add(int a, int b); #endif Most compilers also support #pragma once, which is shorter but not part of the C standard.

A single block is cache friendly and easy to free. int (*grid)[cols] = malloc(rows * sizeof *grid); if (!grid) return -1; grid[2][3] = 5; free(grid); // array of row pointers alternative int **m = malloc(rows * sizeof *m); for (int i = 0; i < rows; i++) m[i] = calloc(cols, sizeof **m); for (int i = 0; i < rows; i++) free(m[i]); free(m); The first form uses a variable length array type.

Use modulo arithmetic and a count. #define CAP 8 typedef struct { int a[CAP]; int front, rear, count; } Queue; int enqueue(Queue *q, int x) { if (q->count == CAP) return -1; q->a[q->rear] = x; q->rear = (q->rear + 1) % CAP; q->count++; return 0; } int dequeue(Queue *q, int *x) { if (q->count == 0) return -1; *x = q->a[q->front]; q->front = (q->front + 1) % CAP; q->count--; return 0; }