Merge pull request neetcode-gh#1035 from ritesh-dt/main

Add solution for 875 - Koko Eating Bananas in C

Author: Ahmad-A0

c/875-Koko-Eating-Bananas.c

@@ -0,0 +1,66 @@
+/*
+    Given a array of bananas piles containing differing amounts of bananas and
+    'h' hours to eat all of them.
+    Determine the minimum speed (i.e. numberbananas per-hour) possible.
+
+    Ex. piles = [3,6,7,11], h = 8 -> 4
+
+    If all the bananas can be eaten with speed 'x' than the same holds true for 
+    any speed more than 'x'. Similarly if bananas cannot be eated at speed 'y', 
+    the same will be speeds less than 'y'. 
+    
+    Binary search can be performed on the possible values of speed till the 
+    minimum speed is found with,
+        left bound = ceil(total/h)
+        right bound = maximum pile size
+
+    Time: O(NlogM) where N is number of piles and M is maximum pile size
+    Space: O(1)
+*/
+
+int hoursRequired(int* piles, int pilesSize, int h, int speed) {
+    int hours = 0;
+    for (int i = 0; i < pilesSize; ++i) {
+        hours += (piles[i]+speed-1)/speed;
+    }
+
+    return hours;
+}
+
+long sumOfArray(int* piles, int pilesSize) {
+    long total = 0l;
+    
+    for (int i = 0; i < pilesSize; ++i) {
+        total += piles[i];
+    }
+    
+    return total;
+}
+
+int maxElement(int* piles, int pilesSize) {
+    int maxElem = piles[0];
+    
+    for (int i = 0; i < pilesSize; ++i) {
+        maxElem = fmax(maxElem, piles[i]);
+    }
+    return maxElem;
+}
+
+int minEatingSpeed(int* piles, int pilesSize, int h){
+    long total = sumOfArray(piles, pilesSize);
+    
+    int l = (total+h-1)/h;
+    int r = maxElement(piles, pilesSize);
+    
+    while (l < r) {
+        int mid = l + (r-l)/2;
+
+        int hours = hoursRequired(piles, pilesSize, h, mid);
+        if (hours <= h)
+            r = mid;
+        else if (hours > h)
+            l = mid + 1;
+    }
+
+    return r;
+}